Problems Solutions Shepherd 52- X- ARNOLD Surveying Problems and Solutions F. A. Shepherd ■£ ■»ii^,^iiA * \ Thfs new book gives a presentation concentrating on mathematical problems, an aspect of the subject which usually causes most difficulty. Summaries of basic theory are followed by worked examples and selected exer- cises. The book covers three main branches of surveying: measurement, surveying techniques and industrial appli- cations. It is a book concerned mainly with engineering surveying as applied, for example, in the construction and mining industries. Contents Linear Measurement Surveying Trigonometry Co-ordinates Instrumental Optics Levelling Traverse Surveys Tacheometry Dip and Fault Problems Areas Volumes Circular Curves Vertical and Transition Curves Values in both imperial and metric (S. units are given in the problems Edward Arnold 80s. net Edward Arnold (Publishers) Ltd., 41 Maddox Street, London, W.I. Printed in Great Britain SURVEYING PROBLEMS & SOLUTIONS Shop l>ord 1 Surveying Problems and Solutions F. A. Shepherd c.En g ., A.R.i.c.s.,M.i.Min.E. Senior Lecturer in Surveying Nottingham Regional College of Technology London. Edward Arnold (Publishers) Ltd. HARRIS Co<...jE | PISTON ! I © F.A. Shepherd 1968 First published 1968 Boards edition SBN: Q7 131 3198 5t Paper edition SBN: 7131 3199 3 GREEK ALPHABET A a alpha N V nu B £ beta B a xi r y gamma o omicron A 8 delta II TT Pi E e epsilon p P rho z c zeta 2 a sigma H V eta T r tau © 6 theta Y V upsilon I i iota $ <f> phi K K kappa X X chi A X lambda W 1> psi M M mu n CO omega Printed in Great Britain by Bookprint Ltd., Crawley, Sussex PREFACE This book is an attempt to deal with the basic mathematical aspects of 'Engineering Surveying', i.e. surveying applied to construction and mining engineering projects, and to give guidance on practical methods of solving the typical problems posed in practice and, in theory, by the various examining bodies. The general approach adopted is to give a theoretical analysis of each topic, followed by worked examples and, finally, selected exer- cises for private study. Little claim is made to new ideas, as the ground covered is elementary and generally well accepted. It is hoped that the mathematics of surveying, which so often causes trouble to beginners, is presented in as clear and readily understood a manner as possible. The main part of the work of the engineering surveyor, civil and mining engineer, and all workers in the construction industry is confined to plane surveying, and this book is similarly restricted. It is hoped that the order of the chapters provides a natural sequence, viz.: (a) Fundamental measurement (i) Linear measurement in the horizontal plane, (ii) Angular measurement and its relationship to linear values, i.e. trigonometry, (iii) Co-ordinates as a graphical and mathematical tool. (b) Fundamental surveying techniques (i) Instrumentation. (ii) Linear measurement in the vertical plane, i.e. levelling, (iii) Traversing as a control system, (iv) Tacheometry as a detail and control system. (c) Industrial applications (i) Three-dimensional aspects involving inclined planes, (ii) Mensuration, (iii) Curve surveying. Basic trigonometry is included, to provide a fundamental mathe- matical tool for the surveyor. It is generally found that there is a deficiency in the student's ability to apply numerical values to trigo- nometrical problems, particularly in the solution of triangles, and it is hoped that the chapter in question shows that more is required than the sine and cosine formulae. Many aspects of surveying, e.g. errors in surveying, curve ranging, etc. require the use of small angles, and the application of radians is suggested. Few numerical problems are posed relating to instrumentation, but it is felt that a knowledge of basic physical properties affords a more complete understanding of the con- struction and use of instruments. To facilitate a real grasp of the sub- ject, the effects of errors are analysed in all sections. This may appear too advanced for students who are not familiar with the element- ary calculus, but it is hoped that the conclusions derived will be beneficial to all. With the introduction of the Metric System in the British Isles and elsewhere, its effect on all aspects of surveying is pin-pointed and conversion factors are given. Some examples are duplicated in the proposed units based on the International System (S.I.) and in order to give a 'feel' for the new system, during the difficult transition period, equivalent S.I. values are given in brackets for a few selected examples. The book is suitable for all students in Universities and Technical Colleges, as well as for supplementary postal tuition, in such courses as Higher National Certificates, Diplomas and Degrees in Surveying, Construction, Architecture, Planning, Estate Management, Civil and Mining Engineering, as well as for professional qualification for the Royal Institution of Chartered Surveyors, the Institution of Civil Engineers, the Incorporated Association of Architects and Surveyors, the Institute of Quantity Surveyors, and the Institute of Building. ACKNOWLEDGMENTS I am greatly indebted to the Mining Qualifications Board (Ministry of Power) and the Controller of H.M. Stationery Office, who have given permission for the reproduction of examination questions. My thanks are also due to the Royal Institution of Chartered Surveyors, the Institution of Civil Engineers, to the Senates of the Universities of London and Nottingham, to the East Midlands Educational Union and the Nottingham Regional College of Technology, all of whom have allowed their examination questions to be used. My special thanks are due to many of my colleagues at Nottingham, but especially to Messrs. J. H. Ball, A.R.I.C.S., A.I.A.S., A.M.I.Min.E., A. Eaton, B.Sc., C.Eng., A.M.I.C.E., A.M.B.I.M., G. M. Lewis, B.Sc, Ph.D., M. B. Pate, M.Sc, A. A. Payne, B.Sc, C. Rayner, B.Sc, A.R.I.C.S., R. Robb, A.R.I.C.S., A.M.I.Min.E., D.B. Shaw, B.Sc, and J. P. Withers, B.Sc, C.Eng., A.M.I.C.E., all of whom have offered advice and help in checking the text The ultimate responsibility for the accuracy is, of course, my own. I am very conscious that, even with the most careful checking, it is not to be expected that every mistake has been eliminated, and I can only ask readers if they will kindly bring any errors to my notice. Nottingham F. A. SHEPHERD 1968 <sr 00 Tf 00 00 co v) "<r Tj- VO 00 00 vo CD CN o 1-4 «-« VO t CN l/> 00 *4 © cp CTv rH rH CO op Y 1—1 V © o O o 1-1 CT\ T-t Oi CO rH © § CN o CN o VO r-i rH CN 00 1-i o o i-4 CO rH rH w CN t> CD lO VO co •<H CN m o i-t § rH O O CN »-H o rH © o o CN VO O CO <J\ 00 © 00 S t> VO o »— 1 o\ o\ t-l "«3- o a o t^ fc i-H O rH 00 o o f^ CO c e* o 00 t>» VO CO CTi o\ J3 ^r o u tH o O 00 o © 00 VO o 10 CO CN o CO cp Cp CO ? 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CO • M co S a 3 Ofl H > ij « CJ ^ £ +j > o CD £ E CO "^3 CD t^ lO O 52 T3 CO ^ « -> © CD if) o £ VO qT n H CS VI CONVERSION FACTORS (B) (Ref Changing to the Metric System, H.M.S.O., 1967) Length 1 mile = 1-609 34 km 1km = 0-621371 mile 1 furlong = 0-201 168 km 1 chain = 20-116 8 m 1yd = 0-914 4 m lm 1-093 61 yd 1ft = 0-304 8 m lin. = 2-54 cm 1 cm = 0-393701 in. 1 fathom = 1-828 8 m llink = 0-201 168 m Area 1 sq. mile = 2-58999 km 2 1 acre = 4046-86 m 2 1 km 2 = 247-105 acres 1 rood = 1011-71 m 2 1yd 2 = 0-836 127 m 2 lm 2 = 1-19599 yd 2 1ft 2 = 0-092903 m 2 lin 2 = 6-4516 cm 2 1 cm 2 = 0-15500 in 2 1 sq. chain = 404-686 m 2 Volume 1yd 3 = 0-764555 m 3 lm 3 = 1-30795 yd 3 1 ft 3 = 0-028 316 8 m 3 lm 3 = 35-314 7 ft 3 lin 3 = 16-387 1 cm 3 1 cm 3 = 0-061023 7 in 3 lgal = 0-004546 09 m 3 = 4-54609 litre 1 litre = 0-2200 gal Velocity 1 mile/h = 1-609 34 km/h 1 km/h = 0-621371 m. p. r lft/s = 0-3048 m/s 1 m/s = 3-28084 ft/s Acceleration 1 ft/s 2 = 0-3048 m/s 2 1 m/s 2 = 3-280 84 ft/s 2 Mass 1 ton = 1016-05 kg 1 cwt = 50-8023 kg lib = 0-453 59237 kg 1kg = 2-20462 lb Vll Mass per unit length 1 lb/ft = 1-488 16 kg/m Mass per unit area lib/ft 2 = 4-88243 kg/m 2 Density 1 ton/yd 3 = 1328-94 kg/m 3 1 lb/ft 3 = 16-018 5 kg/m 3 1 kg/m 3 = 0-062428 lb/ft 3 1 lb/gal = 99-776 3 kg/m 3 0-09978 kg/1 Force Hbf = 4-448 22 N IN = 0-224 809 lbf Ikgf = 9-80665 N 1 kgf = 2-20462 lbf Force (weight) /unit length 1 lbf/ft = 14-593 9 N'm Pressure 1 lbf/ft 2 = 47-880 3 N/m 2 1 N/m 2 = 0-000 145 038 lbf/in 2 1 lbf /in 2 = 6894-76 N/m 2 1 kgf /cm 2 = 98-066 5 kN/m 2 lkgf/m 2 = 9-80665 N/m 2 Standard gravity 32-1740 ft/s 2 = 9-80665 m/s 2 N.B. lib = 0-453 592 kg 1 lbf = 0-453 592 x 9-80665 = 4-448 22 N 1 newton (N) unit of force = that force which applied to a mass of 1 kg gives an acceleration of 1 m/s 2 . Vlll CONTENTS Chapter Page 1 LINEAR MEASUREMENT 1 1.1 The basic principles of surveying 1 1.2 General theory of measurement 2 1.3 Significant figures in measurement and computation 3 1.4 Chain surveying 6 1.41 Corrections to ground measurements 6 1.42 The maximum length of offsets from chain lines 13 1.43 Setting out a right angle by chain 15 1.44 To find the point on the chain line which produces a perpendicular from a point outside the line 16 1.45 Obstacles in chain surveying 17 Exercises 1(a) 22 1.5 Corrections to be applied to measured lengths 23 1.51 Standardisation 23 1.52 Correction for slope 23 1.53 Correction for temperature 26 1.54 Correction for tension 27 1.55 Correction for sag 32 1.56 Reduction to mean sea level 38 1.57 Reduction of ground length to grid length 39 1.6 The effect of errors in linear measurement 45 1.61 Standardisation 45 1.62 Malalignment and deformation of the tape 45 1.63 Reading or marking the tape 46 1.64 Errors due to wrongly recorded temperature 46 1.65 Errors due to variation from the recorded value of tension 47 1.66 Errors from sag 48 1.67 Inaccurate reduction to the horizontal 51 1.68 Errors in reduction from height above or below mean sea level 52 1.69 Errors due to the difference between ground and grid distances 52 Exercises 1(b) 52 2 SURVEYING TRIGONOMETRY 57 2.1 Angular measurement 57 2.11 The degree system 57 2.12 Trigonometrical ratios 58 2.13 Complementary angles 60 2.14 Supplementary angles 60 2.15 Basis of tables of trigonometrical functions 63 2.16 Trigonometric ratios of common angles 64 2.17 Points of the compass 65 2. 18 Easy problems based on the solution of the right- angled triangle 67 Exercises 2(a) 71 2.2 Circular measure 72 2.21 The radian 72 2.22 Small angles and approximations 73 2.3 Trigonometrical ratios of the sums and differences of two angles 77 2.4 Transformation of products and sums 79 2.5 The solution of triangles 80 2.51 Sine rule 80 2.52 Cosine rule 81 2.53 Area of a triangle 82 2.54 Half-angle formulae 82 2.55 Napier's tangent rule 83 2.56 Problems involving the solution of triangles 83 2.6 Heights and distances 91 2.61 To find the height of an object having a vertical face 91 2.62 To find the height of an object when its base is inaccessible 92 2.63 To find the height of an object above the ground when its base and top are visible but not accessible 95 2.64 To find the length of an inclined object on the top of a building 98 2.65 To find the height of an object from three angles of elevation only 100 2.66 The broken base line problem 102 2; 67 To find the relationship between angles in the horizontal and inclined planes 106 Exercises 2(b) 108 CO-ORDINATES 112 3.1 Polar co-ordinates 112 3.11 Plotting to scale 113 3.12 Conversion of the scales 113 3.13 Scales in common use 114 3.14 Plotting accuracy 114 3.15 Incorrect scale problems 114 3.2 Bearings 115 3.21 True north 115 3.22 Magnetic north 115 3.23 Grid north 116 3.24 Arbitrary north 116 3.25 Types of bearing 117 3.26 Conversion of horizontal angles into bearings 121 3.27 Deflection angles 124 Exercises 3(a) 126 3.3 Rectangular co-ordinates 127 3.31 Partial co-ordinates, AE, AN 128 3.32 Total co-ordinates 128 Exercises 3(b) (Plotting) 131 3.4 Computation processes 133 3.41 Computation by logarithms 134 3.42 Computation by machine 134 3.43 Tabulation process 135 3.44 To obtain the bearing and distance between two points given their co-ordinates 136 3.5 To find the co-ordinates of the intersection of two lines 146 3.51 Given their bearings from two known co-ordinate stations 146 3.52 Given the length and bearing of a line AB and all the angles A, B and C 149 Exercises 3(c) (Boundaries) 157 3.6 Transposition of grid 158 3.7 The National Grid Reference system 160 Exercises 3(d) (Co-ordinates) 163 Appendix (Comparison of Scales) 169 4 INSTRUMENTAL OPTICS 170 4.1 Reflection at plane surfaces 170 4.11 Laws of reflection 170 4.12 Deviation by successive reflections on two inclined mirrors 170 4.13 The optical square 171 4.14 Deviation by rotating the mirror 171 4.15 Principles of the sextant 172 4.16 Use of the true horizon 174 4.17 Artificial horizon 175 4.18 Images in plane mirrors 176 4.19 Virtual and real images 177 XI 4.2 Refraction at plane surfaces 177 4.21 Laws of refraction 177 4.22 Total internal reflection 177 4.23 Relationships between refractive indices 178 4.24 Refraction through triangular prisms 179 4.25 Instruments using refraction through prisms 180 Exercises 4(a) 184 4.3 Spherical mirrors 184 4.31 Concave or converging mirrors 184 4.32 Convex or diverging mirrors 186 4.33 The relationship between object and image in curved mirrors 186 4.34 Sign convention lg7 4.35 Derivation of formulae Igg 4.36 Magnification in spherical mirrors 190 4.4 Refraction through thin lenses 191 4.41 Definitions 191 4.42 Formation of images 192 4.43 The relationship between object and image in a thin lens 193 4.44 Derivation of formulae 193 4.45 Magnification in thin lenses 195 4.5 Telescopes 196 4.51 Kepler's astronomical telescope 196 4.52 Galileo's telescope 196 4.53 Eyepieces I97 4.54 The internal focussing telescope 198 4.55 The tacheometric telescope (external focussing) 201 4.56 The anallatic lens 203 4.57 The tacheometric telescope (internal focussing) 207 4.6 Instrumental errors in the theodolite 210 4.61 Eccentricity of the horizontal circle 210 4.62 The line of collimation not perpendicular to the trunnion axis 213 4.63 The trunnion axis not perpendicular to the vertical axis 215 4.64 Vertical axis not truly vertical 217 4-65 Vertical circle index error 219 4.7 The auxiliary telescope 228 4.71 Side telescope 228 4.72 Top telescope 233 Xll 4.8 Angular error due to defective centring of the theodolite 234 4.9 The vernier 237 4.91 Direct reading vernier 237 4.92 Retrograde vernier 238 4.93 Special forms used in vernier theodolites 238 4-94 Geometrical construction of the vernier scale 238 Exercises 4(b) 240 LEVELLING 244 5. 1 Definitions 244 5.2 Principles 245 5.3 Booking, of readings 246 5.31 Method 1, rise and fall 246 5.32 Method 2, height of collimation 247 Exercises 5 (a) (Booking) 254 5.4 Field testing of the level 257 5.41 Reciprocal levelling method 257 5.42 Two-peg method 259 Exercises 5 (b) (Adjustment) 264 5.5 Sensitivity of the bubble tube 267 5.51 Field test 267 5.52 O-E correction 268 5.53 Bubble scale correction 268 Exercises 5(c) (Sensitivity) 270 5.54 Gradient screws (tilting mechanism) 271 5.6 The effect of the earth's curvature and atmospheric refraction 272 5.61 The earth's curvature 272 5-62 Atmospheric refraction 273 5.63 The combined effect of curvature and refraction 273 Exercises 5(d) (Curvature and refraction) 275 5.64 Intervisibility 275 Exercises 5 (e) (Intervisibility) 277 5.65 Trigonometrical levelling 278 5.7 Reciprocal levelling 279 5.71 The use of two instruments 281 Exercises 5(f) (Reciprocal levelling) 282 5.8 Levelling for construction 283 5.81 Grading of constructions 283 Xlll 5.82 The use of sight rails and boning (or travelling) rods 284 5.83 The setting of slope stakes 286 Exercises 5(g) (Construction levelling) 288 Exercises 5 (h) (General) 289 TRAVERSE SURVEYS 298 6. 1 Types of traverse 298 6.11 Open 298 6.12 Closed 298 6.2 Methods of traversing 299 6.21 Compass traversing 300 6.22 Continuous azimuth method 301 6.23 Direction method 302 6. 24 Separate angular measurement 304 Exercises 6(a) 304 6.3 Office tests for locating mistakes in traversing 306 6.31 A mistake in the linear value of one line 306 6.32 A mistake in the angular value at one station 307 6.33 When the traverse is closed on to fixed points and a mistake in the bearing is known to exist 307 6.4 Omitted measurements in closed traverses 308 6.41 Where the bearing of one line is missing 308 6.42 Where the length of one line is missing 309 6-43 Where the length and bearing of a line are missing 309 6.44 Where the bearings of two lines are missing 309 6.45 Where two lengths are missing 314 6.46 Where the length of one line and the bearing of another line are missing 315 Exercises 6(b) (Omitted values) 316 6.5 The adjustment of closed traverses 317 6.51 Where the start and finish of a traverse are fixed 317 6-52 Traverses which return to their starting point 323 6.53 Adjusting the lengths without altering the bearings 323 6.54 Adjustment to the length and bearing 330 6.55 Comparison of methods of adjustment 336 Exercises 6 (c) (Traverse adjustment) 348 Exercises 6(d) (General) 352 XIV 7 TACHEOMETRY 359 7.1 Stadia systems — fixed stadia 359 7.2 Determination of the tacheometric constants m and K 360 7.21 By physical measurement of the instrument 360 7.22 By field measurement 361 7.3 Inclined sights 362 7.31 Staff normal to the line of sight 362 7.32 Staff vertical 363 7-4 The effect of errors in stadia tacheometry 367 7.41 Staff tilted from the normal 367 7.42 Error in the angle of elevation with the staff normal 367 7.43 Staff tilted from the vertical 368 7.44 Accuracy of the vertical angle to conform to the overall accuracy 371 7.45 The effect of the stadia intercept assumption 372 Exercises 7(a) 380 7.5 Subtense systems 383 7.51 Tangential method 383 7.52 Horizontal subtense bar system 388 7.6 Methods used in the field 392 7.61 Serial measurement 392 7.62 Auxiliary base measurement 393 7.63 Central auxiliary base 395 7.64 Auxiliary base perpendicularly bisected by the traverse line 397 7.65 Two auxiliary bases 398 7-66 The auxiliary base used in between two traverse lines 400 Exercises 7(b) 403 8 DIP AND FAULT PROBLEMS 411 8.1 Definitions 411 8.2 Dip problems 413 8.21 Given the rate and direction of full dip, to find the apparent dip in any other direction 413 8.22 Given the direction of full dip and the rate and direction of an apparent dip, to find the rate of full dip 413 8.23 Given the rate and direction of full dip, to find the bearing of an apparent dip 415 XV 8.24 Given two apparent dips, to find the rate and direction of full dip 416 8.25 Given the rate of full dip and the rate and direction of an apparent dip, to find the direction of full dip 421 8.26 Given the levels and relative positions of three points in a plane (bed or seam), to find the direction and rate of full dip 422 8.3 Problems in which the inclinations are expressed as angles and a graphical solution is required 427 8.31 Given the inclination and direction of full dip, to find the rate of apparent dip in a given direction 427 8.32 Given the inclination and direction of full dip, to find the direction of a given apparent dip 428 8.33 Given the inclination and direction of two apparent dips, to find the inclination and direction of full dip 429 Exercises 8(a) 429 8.4 The rate of approach method for convergent lines 432 8.5 Fault problems 437 8.51 Definitions 437 8.52 To find the relationship between the true and apparent bearings of a fault 443 8.53 To find the true bearing of a fault when the throw of the fault opposes the dip of the seam 444 8.54 Given the angle 8 between the full dip of the seam and the true bearing of the fault, to find the bearing of the line of contact 446 8.55 To find the true bearing of a fault when the downthrow of the fault is in the same general direction as the dip of the seam 449 8.56 Given the angle 8 between the full dip of the seam and the true bearing of the fault, to find the bearing of the line of contact 449 8.6 To find the bearing and inclination of the line of intersection (AB) of two inclined planes 450 Exercises 8 (b) (Faults) 452 Exercises 8 (c) (General) 454 AREAS 457 9.1 Areas of regular figures 457 9.11 Areas bounded by straight lines 457 9. 12 Areas involving circular curves 459 9. 13 Areas involving non-circular curves 460 XVI 9.14 Surface areas 461 9.2 Areas of irregular figures 471 9.21 Equalisation of the boundary to give straight lines 471 9.22 The mean ordinate rule 472 9.23 The mid-ordinate rule 473 9.24 The trapezoidal rule 473 9.25 Simpson's rule 474 9.26 The planimeter 477 9.3 Plan areas 481 9.31 Units of area 481 9.32 Conversion of planimetric area in square inches into acres 482 9.33 Calculation of area from co-ordinates 482 9.34 Machine calculations with checks 488 9.4 Subdivisions of areas 490 9.41 The subdivision of an area into specified parts from a point on the boundary 490 9.42 The subdivision of an area by a line of known bearing 491 9.43 The sub-division of an area by a line through a known point inside the figure 492 Exercises 9 497 10 VOLUMES 501 10. 1 Volumes of regular solids 501 10.2 Mineral quantities 509 Exercises 10 (a) (Regular solids) 511 10.3 Earthwork calculations 513 10.31 Calculation of volumes from cross-sectional areas 513 Exercises 10 (b) (Cross- sectional areas) 523 10.32 Alternative formulae for the calculation of volumes from the derived cross-sectional areas 525 10.33 Curvature correction 535 10.34 Derivation of the eccentricity e of the centroid G 537 10.4 Calculation of volumes from contour maps 543 10.5 Calculation of volumes from spot-heights 543 10.6 Mass-haul diagrams 544 10.61 Definitions 544 10.62 Construction of the mass-haul diagram 545 10.63 Characteristics of the mass-haul diagram 546 XV11 10.64 Free-haul and overhaul 546 Exercises 10 (c) (Earthwork volumes) 552 11 CIRCULAR CURVES 559 11.1 Definition 559 11.2 Through chainage 559 11.3 Length of curve L 560 11.4 Geometry of the curve 560 11.5 Special problems 561 11.51 To pass a curve tangential to three given straights 561 11.52 To pass a curve through three points 563 Exercises 11(a) 566 11.53 To pass a curve through a given point P 567 Exercises 11(b) (Curves passing through a given point) 571 11.54 Given a curve joining two tangents, to find the change required in the radius for an assumed change in the tangent length 572 11.6 Location of tangents and curve 575 11.7 Setting out of curves 576 11.71 By linear equipment only 576 11.72 By linear and angular equipment 580 11.73 By angular equipment only 580 Exercises 11(c) 588 11.8 Compound curves 591 Exercises 11(d) (Compound curves; 599 11.9 Reverse curves 600 Exercises 11(e) (Reverse curves) 605 12 VERTICAL AND TRANSITION CURVES 607 12. 1 Vertical curves 607 12.2 Properties of the simple parabola 608 12.3 Properties of the vertical curve 609 12.4 Sight distances 611 12.41 Sight distances for summits 611 12.42 Sight distances for valley curves 613 12.43 Sight distance related to the length of the beam of a vehicle's headlamp 615 12.5 Setting-out data 616 Exercises 12(a) 624 XV111 12.6 Transition curves 627 12.61 Superelevation 627 12.62 Cant 628 12.63 Minimum curvature for standard velocity 628 12.64 Length of transition 629 12.65 Radial acceleration 629 12.7 The ideal transition curve 630 12.8 The clothoid 632 12.81 To find Cartesian co-ordinates 632 12.82 The tangential angle 633 12.83 Amount of shift 633 12.9 The Bernouilli lemniscate 634 12.91 Setting out using the lemniscate 635 12. 10 The cubic parabola 636 12.11 The insertion of transition curves 637 12.12 Setting-out processes 640 12. 13 Transition curves applied to compound curves 644 Exercises 12(b) 649 Abbreviations used for Examination Papers E.M.E.U. East Midlands Educational Union I.C.E. Institution Of Civil Engineers L.U. London University B.Sc. (Civil Engineering) L.U./E London University B.Sc. (Estate Management) M.Q.B./S Mining Qualifications Board (Mining Surveyors) M.Q.B./M Mining Qualifications Board (Colliery Managers) M.Q.B./UM Mining Qualifications Board (Colliery Undermanagers) N.R.C.T. Nottingham Regional College of Technology N.U. Nottingham University R.I.C.S./G Royal Institution of Chartered Surveyors (General) R.I.C.S./M Royal Institution of Chartered Surveyors (Mining) R.I.C.S./ML Royal Institution of Chartered Surveyors (Mining/Land) R.I.C.S./Q Royal Institution of Chartered Surveyors (Quantity) LINEAR MEASUREMENT 1.1 The Basic Principles of Surveying Fundamental rule 'Always work from the whole to the part*. This implies 'precise control surveying' as the first consideration, followed by 'subsidiary detail surveying'. A point C in a plane may be fixed relative to a given line AB in one of the following ways: 1. Triangulation Angular measurement from a fixed base line. The length AB is known. The angles a and /3 are measured. »£ B a. Xe li .V Fig. 1.1(a) 2. Trilateration Linear measurement only. The lengths AC and BC are measured or plotted. The position of C is always fixed provid- ed AC + BC > AB. Uses: (a) Replacing triangulation with the use of microwave mea- suring equipment. (b) Chain surveying. A Bt Fig. 1.1(b) 1 SURVEYING PROBLEMS AND SOLUTIONS 3. Polar co-ordinates Linear and angular measurement. Uses: (a) Traversing. (b) Setting out. (c) Plotting by protractor. ,-° c (s,6) BhT Fig. 1.1(c) 4. Rectangular co-ordinates Linear measurement only at right-angles. Uses: (a) Offsets. (b) Setting out. (c) Plotting. A A Bit 90" ■OC Fig. 1.1(d) 1.2 General Theory of Measurement The following points should be noted: (1) There is no such thing as an exact measurement. All measure- ments contain some error, the magnitude of the error being dependent on the instruments used and the ability of the observer. (2) As the true value is never known, the true error is never deter- LINEAR MEASUREMENT 3 mined. (3) The degree of accuracy, or its precision, can only be quoted as a relative accuracy, i.e. the estimated error is quoted as a fraction of the measured quantity. Thus 100 ft measured with an estimated error of 1 inch represents a relative accuracy of 1/1200. An error of lcm in 100 m = 1/10000. (4) Where readings are taken on a graduated scale to the nearest subdivision, the maximum error in estimation will be ± l / 2 division. (5) Repeated measurement increases the accuracy by y/n, where n is the number of repetitions. N.B. This cannot be applied indefinite- ly- (6) Agreement between repeated measurements does not imply accuracy but only consistency. 1.3 Significant Figures in Measurement and Computation If a measurement is recorded as 205 ft to the nearest foot, its most probable value is 205 ±0*5 ft, whilst if measured to the nearest 0*1 ft its most probable value is 205-0 ± 0-05 ft. Thus the smallest recorded digit is subject to a maximum error of half its value. In computation, figures are rounded off to the required degree of precision, generally by increasing the last significant figure by 1 if the following figure is 5 or more. (An alternative is the rounding off with 5 to the nearest even number.) Thus 205-613 becomes 205-61 to 2 places, whilst 205-615 becomes 205-62 to 2 places, or 205-625 "may also be 205*62, giving a less biased value. It is generally better to work to 1 place of decimals more than is required in the final answer, and to carry out the rounding-off process at the end. In multiplication the number of significant figures depends on the accuracy of the individual components, e.g., if P = x.y, then P + 8P = (x + 8x)(y + 8y) = xy + x8y + y8x + 8x8y Neglecting the last term and substracting P from both sides of the equation, 8p = x8y + ySx ~ P gives S£ = ^§X + y8x = 8y + 8x P xy xy sp - p (f + t) (11) 4 SURVEYING PROBLEMS AND SOLUTIONS Thus the relative accuracy of the product is the sum of all the relative accuracies involved in the product. Example 1.1 A rectangle measures 3-82 in. and 7-64 in. with errors of ± 0*005 in. Express the area to the correct number of significant figures. P = 3-82 x 7-64 = 29*184 8 in 2 relative accuracies ° ~ _i_ 3-82 ~ 750 0-005 .. 1 7-64 1500 500 SP = 290- + -L-) = ™ \750 1500/ = ± 0-06 .-. the area should be given as 29-2in 2 . As a general rule the number of significant figures in the product should be at least the same as, or preferably have one more significant figure than, the least significant factor. The area would thus be quoted as 29-18 in 2 In division the same rule applies. Q = - y Q + 8Q = x + 8x = * + £f - rf^ + ... y + 8y y y y 2 Subtracting Q from both sides and dividing by Q gives SQ = Q (?I - *) (1.2) Powers R = x n R + 8R = (x + 8x) n = x n + n8x + ... 8R n8x . .. , — = — — i.e. nx relative accuracy of single value. 8R = n8x (1 . 3) Roots This is the opposite relationship R = ^x .'. R n = x From the above R n + n8R = x + 8x LINEAR MEASUREMENT 5 nSR = Sx 8R _ 8x_ R n ~ nx 8R = -8x (1.4) Example 1.2 If R = (5-01 ± 0-005) 2 5-01 2 = 25-1001 8R = 2 x 0-005 = 0-01 .'. R should be given as 25*10 Example 1.3 If R = V 25 * 10 ± °* 01 v'25-10 = 5-009 9 8R = ^ = 0-005 .*. R should be given as 5-01 Example 1.4 A rectangular building has sides approximately 480 metres and 300 metres. If the area is to be determined to the nearest 10 m 2 what will be the maximum error permitted in each line, assuming equal precision ratios for each length? To what degree of accuracy should the lines be measured? A = 480 x 300 - 144 000 m 2 8A = 10 m 2 8A = _1 = §x Sy A 14400 x + y but 8x = 8y . 8x 8y _ 28x x y x y ~ x 8x_ = 1 = 1 x 2 x 14400 28 800 i.e. the precision ratio of each line is *„ 28 800 This represents a maximum in 480 m of - = 0*016 7 m Zq 800 and in 300 m of -i9p- = 0-0104 m 2o 800 If the number of significant figures in the area is 5, i.e. to the nearest 10 m 2 , then each line also must be measured to at least 5 sig- nificant figures, i.e. 480-00 m and 300-00m. SURVEYING PROBLEMS AND SOLUTIONS 1.4 Chain Surveying The chain There are two types : (a) Gunter's chain 1 chain* = 100 links = 66 ft 1 link = 0-66 ft = 7-92 in. Its advantage lies in its relationship to the acre 10 sq chains = 100 000 sq links = 1 acre. (b) Engineer's chain 100 links = 100 ft (Metric chain 100 links = 20 m 1 link = 0-2 m) Basic figures There are many combinations of chain lines all dependent on the linear dimensions forming trilateration, Fig. 1 .2. Tie line C A Tie lines Fig. 1 .2 Basic figures in chain surveying 1.41 Corrections to the ground measurements Standardisation Where the length of the chain or tape does not agree with its nom- * See conversion factors, pp. v — vii. LINEAR MEASUREMENT 7 inal value, a correction must be made to the recorded value of a meas- ured quantity. The following rules apply : (1) If the tape is too long, the measurement will be too short — the correction will be positive. (2) If the tape is too short, the measurement will be too long — the correction will be negative. If the length of tape of nominal length / is / ± 81, fil the error per unit length = ± — If the measured length is d m and the true length is d t , then d t = d m ± d m — = ^(l±f) d-5) Alternatively, 1 + 81 actual length of tape (1.6) d m I nominal length of tape d > = *» j 1 ± t) (1 - 5 > Example 1.5 A chain of nominal length 100 links, when compared with a standard, measures 101 links. If this chain is used to measure a line AB and the recorded measurement is 653 links, what is the true length AB? Error per link = -i— = 0*01 100 .'. true length = 653(1 + 0-01) = 653 + 6-53 = 659-53 links . Alternatively, true length = 653 x ^ = 659-53 links . Effect of standardisation on areas Based on the principle of similar figures, true a,ea (.,) . apparent area (,„) x (,££££ 5%. )' SURVEYING PROBLEMS AND SOLUTIONS or A T = A M (l ±y) (1.8) Effect of standardisation on volumes Based on the principle of similar volumes, , / true length of tape V true volume V T = apparent volume x ( apparent length of tap J ue. V r = V„(l ±^)° (110) N.B. Where the error in standardisation is small compared to the size of the area, the % error in area is approximately 2 x % error in length. Example 1.6 A chain is found to be 0*8 link too long and on using it an area of 100 acres is computed. ™. . inn A00-8 \ 2 The true area = 1UU I - TqTT) = 100 x 1-008 2 = 101-61 acres alternatively, linear error = 0*8% ••■ area error = 2 x 0*8 = 1*6% acreage = 100 + 1*6 acres = 101*6 acres This is derived from the binomial expansion of (1 + x) z = 1 + 2x + x 2 i.e .if x is small x z may be neglected /. (1 + x) 2 a 1 + 2x Correction for slope (Fig. 1.3) This may be based on (1) the angle of inclination, (2) the difference in level between the ends of the line. Fig. 1.3 (page 9) Length AC measured (/) Horizontal length AB required (h) Difference in level between A and C (d) Angle of inclination (a) Correction to measured length (c) LINEAR MEASUREMENT h Fig. 1.3 (1) Given the angle of inclination a AB = AC cos a i.e. h = / cos a (1.11) c = I - h = I - I cos a = /(1-cosa) = / versine a (1-12) N.B. The latter equation is a better computation process. Example 1.7 If AC = 126-3 m, a = 2°34\ byEq.(l.ll) AB = 126-3 cos 2°34' = 126-3 x 0-999 = 126-174 m or by Eq. (1.12) c = 126-3 (1 - 0-999) = 126-3 x 0-001 = 0-126 m AB - 126-3 - 0-126 = 126-174 m Example 1.8 In chaining, account should be taken of any significant effect of the slope of the ground on the accuracy of the horizontal length. Calculate the minimum angle of inclination that gives rise to relative accuracies of 1/1000 and 1/3000. From Eq. (1.12), c = I - h = 1(1 - cos a) c = _J_ T 1000 If 1 - cos a 10 SURVEYING PROBLEMS AND SOLUTIONS cos a = 1 - 0-001 = 0-999 a = 2°34' (i.e. 1 in 22) Also, if -j = = = 1 - cos a 3000 cos a = 1 - 0-00033 = 0-99967 a = 1°29' (i.e. 1 in 39) If the difference in level , d, is known h = (I 2 - d 2 y = j(/-d)x (/+ d)}* or I 2 = h 2 + d 2 = (/ - cf + d 2 = I 2 - 2lc + c 2 + d 2 .-. c 2 - 2lc = -d 2 c(c-2l) = -d 2 c = -d 2 c-2l c ~ d z — as c is small compared (1.13) Rigorously, using the binomial expansion, c - I - (I 2 - d 2 y -'-<-£)' d 2 d' = Tl ~ gji ♦ ••• (1.15) The use of the first term only gives the following relative accura- cies (the units may be ft or metres). Gradient Error per 100 ft or m Relative accuracy lin4 0-051 ft (or m) 1/2000 1 in 8 0-003 1 ft (or m) 1/30 000 1 in 10 0-001 3 ft (or m) 1/80000 1 in 20 0-000 1 ft (or m) 1/1 000 000 Thus the approximation is acceptable for: Chain surveying under all general conditions. Traversing, gradients up to 1 in 10. Precise measurement (e.g. base lines), gradients up to 1 in 20. LINEAR MEASUREMENT 11 For setting out purposes Here the horizontal length (h) is given and the slope length (/) is required. / = h sec a c = h sec a - h = h(sec a - 1) (1.16) Writing sec a as a series 1 + ^- + ^+ •••, where a is in radians, i „ x see p. 72. ho. 2 — -s- («• in radians) (1.17) ~ ^(0-017 45a) 2 2i 1-53 ft x 10~ 4 x a 2 (a in degrees) (1.18) - 1*53 x 10" 2 x a 2 per 100 ft (or m) (1.19) Example 1.9 If h = 100 ft (orm), a = 5°," by Eq. (1.16) c = 100(1-003 820-1) = 0-382 Oft (orm) per 100 ft (orm) or by Eq. (1.18) c = 1-53 x 100 x 10" 4 x 5 2 = 1-53 x 25 x 10~ 2 = 0-382 5 ft (or m) per 100 ft (or m) Correction per 100ft (orm) 1° 0-015 ft (orm) 6° 0-551 ft (orm) 2° 0-061 ft (orm) 7° 0-751 ft (orm) 3° 0-137 ft (orm) 8° 0-983 ft (orm) 4° 0-244 ft (orm) 9° 1-247 ft (orm) 5° 0-382 ft (orm) 10° 1-543 ft (orm) If the difference in level, d, is given, I 2 = h 2 + d 2 (h + cf = h 2 + d 2 h 2 + 2hc + c 2 = h 2 + d 2 c(2h + c) = d 2 c- -*- 2h + c 12 SURVEYING PROBLEMS AND SOLUTIONS 2h or rigorously 2h + 8h 3 (1.20) (1.21) N.B. If the gradient of the ground is known as 1 vertical to n horizon- tal the angle of inclination a ~ S (1 rad ~ 57" 3°) n e.g. 1 in 10 gives 57 10 5-7 c To find the horizontal length h given the gradient 1 in n and the measured length I n\fn z + 1 / V« 2 + i n 2 + 1 h -. lny/n z + 1 « 2 +l (1.22) As an alternative to the above, h = I - c - '"IT but if the gradient is given as 1 in n, then d c* — n Fig. 1.4 h ^ I 2nH \ 2n 2 J This is only applicable where n > 20. (1.23) Example 1.10 If a length of 300 ft (orm) is measured on a slope of 1 in 3, the horizontal length is given as: by Eq. (1.22) h = 300x3^10 = 9Q x 3 . 1623 10 - 284-61 ft (orm) LINEAR MEASUREMENT To find the inclined length I given the horizontal length h and the gradient (1 in n) 13 ± = V" 2 + 1 h n , __ hy/n z + 1 (1-24) Example 1.11 If h = 300 ft (orra) and the gradient is 1 in 6, by Eq. (1.24) / = 3Q0 ^ = 50 x 6-083 6 = 304 -15 ft (or m) 1.42 The maximum length of offsets from chain lines A point P is measured from a chain line ABC in such a way that B, P is measured instead of BP, due to an error a in estimating the perpendicular, Fig. 1.5. Fig. 1.5 On plotting, P, is fixed from B, . Thus the displacement on the plan due to the error in direction a PP, = B,P a(radians) _ /a, = 206265 (N.B. 1 radian = 206265 seconds of arc) 14 SURVEYING PROBLEMS AND SOLUTIONS If the maximum length PP % represents the minimum plotable point, 12 i.e. 0*01 in which represents ^— xft, where x is the representative fraction 1/x, then 0-000 83 x = la I 206 265 171-82 x a Assuming the maximum error a = 4°, i.e. 14400 , = 171-82* o-012 x (1.25) 14400 If the scale is 1/2500, then x = 2500, and / = 2500 x 0-012 = 30ft (^10 m) If the point P lies on a fence approximately parallel to ABC, Fig. 1.6, then the plotted point will be in error by an amount P^P 2 = 1(1- cos a). (Fig. 1.5). Boundary line 12 (1 - cos a) Example 1.12 If a = 4°, by Eq. 1.26 0-01 x fl C Fig. 1 .6 , = 0-01 x (1>26) / 12 x (1-0-9976) 0-35 x (1.27) Thus, if x= 2500, / = 875 ft (267 m) The error due to this source is almost negligible and the offset is only limited by practical considerations, e.g. the length of the tape. It is thus apparent that in fixing the position of a point that is critical, e.g. the corner of a building, the length of a perpendicular off- set is limited to 0*012 x ft, and beyond this length tie lines are required, LINEAR MEASUREMENT the direction of the measurement being ignored, Fig. 1.7. 15 Fig. 1.7 1.43 Setting out a right angle by chain From a point on the chain line (Fig. 1.8) (a) (i) Measure off BA = BC (ii) From A and C measure off AD = CD (Proof: triangles ADB and DCB are congruent, thus ABD = DBC = 90 c as ABC is a straight line) Fig. 1.8 Z / / / 7k \ \ \ 8 B Fig. 1.9 (b) Using the principle of Pythagoras, z z = x z + y 2 (Fig. 1.9) By choosing suitable values the right angle may be set out. The basic relationship is x:y:z :: 2n + l : 2n(rc + l) : 2n(n + l)+l. (1.28) If n = I, 2n + l=3 2n(n + 1) = 4 2n(n + 1) + 1 = 5. Check: \2n(n + 1) + l} 2 = (2n 2 + 2n + l) 2 (2n + if + \2n(n + 1)| 2 = 4n 2 + 4n + 1 + 4n 4 + 8n 3 + 4n 2 = 4n z + 8n 3 + 8n 2 + 4n + 1 16 SURVEYING PROBLEMS AND SOLUTIONS = (2n 2 + 2n + If. Check: 5 2 - 3 2 + 4 2 or 25 - 9 + 16. Similarly, if n = 3/4, In + 1 = 6 , 10 4 4 = 40 16 2n(rc + 1) = 6/3 , \ 6 7 4U +1 j = 4 X 4 = 42 16 2n(n + 1) + 1 = 16 16 16 58 16 Thus the ratios become 40 : 42 : 58 and this is probably the best combination for 100 unit measuring equipment; e.g. on the line ABC, Fig. 1.10, set out BC = 40 units. Then holding the ends of the chain at B and C the position of D is fixed by pulling taut at the 42/58 on the chain. Alternative values for n give the following: n = 2 5, 12, 13 (Probably the best ratio for 30 m tapes) n = 3 7, 24, 25 n = 4 9, 40, 41. 1.44 To find the point on the chain line which produces a perpendic- ular from a point outside the line (1) When the point is accessible (Fig. 1.11). From the point D swing the chain of length > DB to cut the chain line at a and b. The required position B is then the mid-point of ab. LINEAR MEASUREMENT 17 (2) When the point is not accessible (Fig. 1.12). From D set out lines Da and Db and, from these lines, perpendicular ad and be. The inter- section of these lines at X gives the the line DX which when produced gives B, the required point. To set out a line through a given point parallel to the given chain line (Fig. 1.13). Given the chain line AB and the given point C. From the given point C bisect the line CB at X. Measure AX and produce the line to D such that AX = XD. CD will then be parallel to AB. 1.45 Obstacles in chain surveying 18 SURVEYING PROBLEMS AND SOLUTIONS (1) Obstacles to ranging (a) Visibility from intermediates (Fig. 1.14). Required to line C and D on the line AB. Place ranging pole at d, and line in c, on line Ad } . From B ob- serve c, and move d 2 on to line Be, . Repetition will produce c 2 , c 3 and d 2 , d 3 etc until C and D lie on the line AB. (b) Non-visibility from intermediates (Fig. 1.15). Required to measure a long line AB in which A and B are not inter- visible and intermediates on these lines are not possible. Set out a 'random line' AC approximately on the line AB. From B find the perpendicular BC to line AC as above. Measure AC and BC. Calculate AB. (2) Obstacles to chaining (a) No obstacle to ranging (i) Obstacle can be chained around. There are many possible variations depending on whether a right angle is set out or not. o*= A B Fig. 1.15 A6 (m) Fig. 1.16 By setting out right angles I Set out equal perpendiculars Bb and Cc; then be - BC. II Set out Bb. Measure Bb and bC. Compute BC. HI Set out line Bb. At b set out the right angle to give C on the chain line. Measure Bb and bC. Compute BC. IV and V Set out parallel lines be as described above to give similar figures, triangles BCX and bcX. Then BC = be x BX bX (1.29) LINEAR MEASUREMENT 19 Fig. 1.17 VI Set out line be so that bB = Be. Compute BC thus, BC* = CO 2 Be + jCcf bB _ be but bB = Be, . RC z BbjbC 2 - Cc 2 ) _. DK - ^ = 2fib -BbxBb = UbC z + Cc 2 ) - Bb 2 (1.30) (1.31) Proof. In Fig. 1.18 using the cosine rule (assuming 6 > 90°), see p. 81 ,2 _ v 2 x 2 + d 2 + 2xd cos0 and ? 2 = y 2 + d 2 - 2yd cosO 2d cos 6 = p - JT _ y2 +d 2_ ? 2 P 2 y - * 2 y - d 2 y = xy 2 + d 2 x - q 2 x d 2 (x + y) = <? 2 x + p 2 y - xy(x + y) 20 SURVEYING PROBLEMS AND SOLUTIONS d 2 = q z x + p 2 y x + y - xy --/ Q x + p y x + y - xy (1.32) If x = y, (ii) Obstacle cannot be chained around. A river or stream represents . this type of obstacle. Again there are many variations depending on whether a right angle is set out or not. By setting out right angles (Fig. 1.19). A random line DA^ is set out and from perpendiculars at C and B points C and B are obtained. By similar triangles DC,C and C 1 B 1 B 2 , DC CB DC CC X Bfi, - CC y CB x CC, BB. - CC, Fig. 1.20 LINEAR MEASUREMENT 21 Without setting out a right angle (Fig. 1.20). A point F is chosen. From points B and C on line AE, BF and CF are measured and produced to G and H. BF = FG and CF = FH. The intersection of DF and GH produce to intersect at J. Then HJ = CD. (iii) Obstacles which obstruct ranging and chaining. The obstruction, e.g. a building, prevents the line from being ranged and thus produced beyond the obstacle. By setting out right angles (Fig. 1 .21) On line ABC right angles are set out at B and C to produce B and C, , where SB, = CC, . ' B, C, is now produced to give D, and E t where right angles are set out to give D and E, where D^D=E^E=BB y = CC, . D and E are thus on the line ABC produced and 0,0, = DC. A A Fi B-l.21 Fig. 1.22 Without setting out right angles (Fig. 1.22) On line ABC, CB is measured and G set out to form an equilateral triangle, i.e. CB= CG = BG. BG is produced to J. An equilateral triangle HKJ sets out the line JE such that JE = BJ. A further equilateral triangle ELD will restore the line ABC pro- duced . The missing length BE = BJ = EJ . 22 SURVEYING PROBLEMS AND SOLUTIONS Exercises 1 (a) 1. The following measurements were made on inclined ground. Reduce the slope distances to the horizontal giving the answer in feet. (a) 200-1 yd at 1 in 2^ (b) 485*5 links at 1 in 5-75 (c) 1/24 th of a mile at 1 in 10-25 (Ans. (a) 557-4 ft (b) 315-7 ft (c) 218-9 ft) 2. Calculate the acreage of an area of 4 in 2 on each of the plans drawn to scale, 2 chains to 1 in., 1/63 360, 1/2500 and 6 in. to 1 mile respectively. (Ans. 1-6, 2560, 3-986, 71-1 acres) 3. A field was measured with a chain 0-3 of a link too long. The area thus found was 30 acres. What is the true area? (I.C.E. Ans. 30-18 acres) 4. State in acres and decimals thereof the area of an enclosure mea- suring 4 in. square on each of three plans drawn to scale of 1/1584, 1/2500, 1/10560 respectively. (Ans. 6-4, 15-9, 284-4 acres) 5. A survey line was measured on sloping ground and recorded as 386-6 ft (117-84 m). The difference of elevation between the ends was 19-3 ft (5-88 m). The tape used was later found to be 100-6 ft (30-66 m) when com- pared with a standard of 100 ft (30-48 m). Calculate the corrected horizontal length of the line. (Ans. 388-4 ft (118-38 m)) 6. A plot of land in the form of a rectangle in which the length is twice the width has an area of 180000 ft 2 . Calculate the length of the sides as drawn on plans of the follow- ing scales. (a) 2 chains to 1 inch, (b) 1/25000. (c) 6 inch to 1 mile. (Ans. (a) 4-55 x 2-27 in. (b) 0-29 x 0-14 in. (c) 0-68 x 0-34 in.) 7. (a) Express the following gradients in degrees to the horizontal: 1 in 3, 1 in 200, 1 in 0-5, being vertical to horizontal in each case. (b) Express the following scales as fractions: 6 in. to 1 mile, 1 in. to 1 mile, 1 in. to 1 chain, 1/8 in. to 1 ft. (c) Express the following scales as inches to 1 mile: 1/2500, 1/500, 1/1080. (M.Q.B./UM Ans. (a) 18°26', 0°17', 63°26' (b) 1/10560, 1/63360, 1/792, 1/96 (c) 25-34, 126-72, 58-67) LINEAR MEASUREMENT 23 8. Find, without using tables, the horizontal length in feet of a line recorded as 247*4 links when measured (a) On ground sloping 1 in 4, (b) on ground sloping at 18°26' (tanl8°26' = 0-333). (Ans. (a) 158-40 (b) 154-89 ft) 9. Show that for small angles of slope the difference between the horizontal and sloping lengths is h z /2l (where h is the difference of vertical height of the two ends of a line of sloping length /). If errors in chaining are not to exceed 1 part in 1000, what is the greatest slope that can be ignored? (L.U./E Ans. 1 in 22*4) 1.5 Corrections to be Applied to Measured Lengths For every linear measurement the following corrections must be considered, the need for their application depending on the accuracy required. 1. In all cases (a) Standardisation. (b) Slope. 2. For relative accuracies of 1/5000 plus (a) Temperature. (b) Tension, (c) Sag. (where applicable) 3. For special cases, 1/50000 plus (a) Reduction to mean sea level. (b) Reduction to grid. Consideration has already been given, p. 6/9, to both standardisa- tion and reduction to the horizontal as they apply to chain surveying but more care must be exercised in precise measurement reduction. 1.51 Standardisation The measuring band in the form of a tape or wire must be compared with a standard under specified conditions of temperature (t s ) and ten- sion (T s ) . If there is any variation from the nominal length then a stan- dardisation correction is needed as already shown. A combination of temperature and standardisation can be seen under correction for tem- perature. 1.52 Correction for slope Where the inclination of the measured length is obtained by mea- surement of the vertical angle the following modification should be noted. Let the height of the instrument be ft, the height of the target h 2 24 SURVEYING PROBLEMS AND SOLUTIONS the measured vertical angle 6 the slope of the measured line a the length of the measured line / Fig. 1.23 In Fig. 1.23, a = d + 8$. In triangle A A B Z B^ by the sine rule (see p. 80), (h, -h z ) sin (90 + 0) sin 86 86" I (fc, - h 2 ) cos 6 / 206 265 (h } -h z ) cos 6 _ (1.34) (1.35) N.B. The sign of the correction conforms precisely to the equation. (1) If /i i = h z , 86 = a = 6 (2) If h\ < h Z z and 6 is +ve, 86 is -ve (Fig. 1.24a) (3) If h,> h z and 6 is -ve, 86 is -ve (Fig. 1.24d) if a is +ve, 86 is +ve (Fig. 1.24b) (4) If /i, < h z and 6 is -ve, 86 is +ve (Fig. 1.24c) Example 1.13 If ^ = 4-5ft(l-37m), h z = 5-5ft (1.68m), 6 = +4°30' / = 350 ft (106-68 m) 206 265 (4-5 - 5-5) cos4°30' then 86 = 350 - -588" = -0°09'48" a = +4 o 30'00" - 0°09'48" = +4° 20 '12" LINEAR MEASUREMENT 25 /> 4 -/>, (a) (b) (c) (d) Fig. 1.24 Correction to measured length (by Eq. 1.12), c = -1(1 -cos0°) = -350(1 -cos 4° 20' 12") = -350(1-0-99714) = -350 x 0-00286 = - 1-001 ft (0-3051 m) ••• Horizontal length = 348-999 ft (106-3749 m) If the effect was ignored; Horizontal length = 350 cos 4° 30' = 348-922 ft (106-3514 m) ••• Error = 0-077 ft (0-0235 m) 26 SURVEYING PROBLEMS AND SOLUTIONS 1.53 Correction for temperature The measuring band is standardised at a given temperature (t 8 ). If in the field the temperature of the band is recorded as (* m ) then the band will expand or contract and a correction to the measured length is given as c = la(t m - t a ) (1.36) where / = the measured length a = the coefficient of linear expansion of the band metal. The coefficient of linear expansion (a) of a solid is defined as 'the increase in length per unit length of the solid when its temper- ature changes by one degree'. For steel the average value of a is given as 6-2 x 10" 6 per °F Since a change of 1 °F = a change of 5/9 °C, using the value above gives a = 6-2 x 10" 6 per 5/9 °C = 11-2 x 10" 6 per °C The range of linear coefficients a is thus given as : Steel Invar per 1°C -6 10-6 to 12-2 (x 10 ) 5-4 to 7-2 (xl0~ 7 ) per 1°F 5-9 to 6-8 3 to 4 To find the new standard temperature t' 8 which will produce the nominal length of the band. Standard length at t a = I ± 81 To reduce the length by 5/: 81 = (/ ± 5/).a.r where t = number of degrees of temperature change required 81 t = (I ±8t)a ' ; - »• * orkz (137 > As 81 is small compared with /, for practical purposes 81 la K = t s ± £f (1.38) LINEAR MEASUREMENT 27 Example 1.14 A traverse line is 500 ft (152*4 m) long. If the tape used in the field is 100 ft (30-48 ra) when standardised at 63 °F (17-2 °C), what correction must be applied if the temperature at the time of mea- surement is 73 °F (22-8 °C)? (Assume a = 6-2 x 10~ 6 per deg F = 11-2 x 10~ 6 per deg C) From Eq. (1.36) c m = 500 x 6-2 x 10~ 6 x (73 - 63) = +0-031 Oft or c (m) = 152-4 x 11-2 x 10" 6 x (22-8 - 17-2) = +0-009 6 m Example 1.15 If a field tape when standardised at 63 °F measures 100-005 2 ft, at what temperature will it be exactly the nominal value? (Assume a = 6-5 x 10" 6 per deg F) SI = +0-0052 ft •'. from Eq. (1.37) t' =63 0-0052 100 x 6-5 x 10" 6 = 63°F-8°F = 55 °F In its metric form the above problem becomes: If a field tape when standardised at 17-2 °C measures 100-005 2 m, at what temperature will it be exactly the nominal value? (Assume a = 11-2 x 10" 6 per deg C) 81 = +0-005 2 m .'. from Eq.(1.37) t' s = 17#2 _ 0-0052 100 x 11-2 x 10" 6 = 17-2 °C - 4-6 °C = 12-6 °C (=54-7°F) 1.54 Correction for tension The measuring band is standardised at a given tension (T s ). If in the field the applied tension is (T m ) then the tape will, due to its own elasticity, expand or contract in accordance with Hooke's Law. A correction factor is thus given as L(T m - T s ) c = A _ E (1.39) 28 SURVEYING PROBLEMS AND SOLUTIONS where L = the measured length (the value of c is in the same unit as L), A = cross-sectional area of the tape, E = Young's modulus of elasticity i.e. stress/strain. The units used for T, A and E must be compatible, e.g. T(lbf) A (in 2 ) E (lbf/in 2 ) or T,(kgf) 4, (cm 2 ) E, (kgf/cm 2 ) (metric) or T 2 (N) A,(m 2 ) E 2 (N/m 2 ) (new S.I. units) Conversion factors lib = 0-453592 kg 1 in 2 = 6-451 6 x 10~ 4 m 2 .'. lib/in 2 = 703-070 kg/m 2 Based on the proposed use of the International System of Units (S.I. units) the unit of force is the Newton (N), i.e. the force required to accelerate a mass of 1 kg 1 metre per second per second . The force 1 lbf = mass x gravitational acceleration = 0-453592 x 9-80665 m/s 2 (assuming standard value) = 4-448 22N lkgf = 9-806 65 N (1 kg = 2-204 62 lb) whilst for stress 1 lbf/in 2 = 6894-76 N/m 2 For steel, E ~ 28 to 30 x 10 6 lbf/in 2 (British units) ~ 20 to 22 x 10 5 kgf/cm 2 (Metric units) ot 19-3 to 20-7 x 10 10 N/m 2 (S.I. units) For invar, E ~ 20 to 22 x 10 6 lbf/in 2 ^ 14 to 15-5 x 10 5 kgf/cm 2 ~ 13-8 to 15*2 x 10 ,0 N/m 2 N.B. (1) If T m = T s no correction is necessary. (2) It is generally considered good practice to over tension to minimise deformation of the tape, the amount of tension being strictly recorded and the correction applied. (3) The cross-sectional area of the tape may be physically measured using a mechanical micrometer, or it may be com- puted from the total weight W of the tape of length L and a value p for the density of the material. W A - j- (1.40) LINEAR MEASUREMENT 29 Example 1.16 A tape is 100 ft at a standard tension of 251bf and mea- sures in cross-section 0-125 in. x 0-05 in. If the applied tension is 20 lbf and E = 30 x 10 5 lbf/in 2 , calculate the correction to be applied. Rv F„1C.Q . 10 ° X (20 - 25) By t-q. l.iy c = . i = -0*009 7 ft (0-125 x 0-05) x (30 x 10 6 ) ° °° 27 " Converting the above units to the metric equivalents gives c = 30-48 m x (9-072 - 11-340) kgf (40-32 x 10~ 7 )m 2 x (21-09 x 10 9 ) kgf/m 2 = -0-008 13 m (i.e. -0-002 7 ft) Based on the International System of Units, 2-268 kgf = 2-268 x 9-806 65 N = 22-241 N or 5 lbf = 5 x 4-448 22 N = 22-241 N. For stress, (21-09 x 10 10 ) kgf/m 2 = 21-09 x 10 9 x 9-80665 = 20-684 x 10 ,o N/m 2 or (30 x 10 6 ) lbf/in 2 = 30 x 10 6 x 6894-76 = 20-684 x 10'° N/m 2 Thus, in S.I. units, 30-48 x 22-241 N c = (40-32 x 10 -7 ) x (20-684 x 10 ,o )N/m 2 = - 0-008 13 m Measurement in the vertical plane Where a metal tape is freely suspended it will elongate due to the applied tension produced by its own weight. The tension is not uniform and the stress varies along its length. Given an unstretched tape AB and a stretched tape AB, , Fig. 1.25, let P and Q be two close points on the tape which become ^Q, under ' tension. If AP = x, AP t = x + s, where s is the amount of elongation of AP . Let PQ = dx, then P, Q, = dx + ds and the strain in %Q,= — as ds is the increase in length and dx is the original length. If T is the tension at P, , in a tape of cross-section A, and E is Young's modulus, then T = EA*1 m dx (!) Given that the load per unit length at P, is w then in P, Q, the load = w.dx being the difference in tension between P, and Q, 30 SURVEYING PROBLEMS AND SOLUTIONS if the tension at Q is T + dT T-(T + dT) = wdx i.e. dT = -wdx (2) x + s dx+ds Fig. 1.2 5 Elongation in a suspended tape In practice the value w is a function of x and by integrating the two equations the tension and extension are derived. Assuming the weight per unit length of the tape is w with a sus- pended weight W, then from (2) dT = -wdx T = -wx + c (3) and (1) T = EA^. dx ds EA -t— = -wx + c dx EAs = iwx 2 + ex + d (4) When T = W, x= I and when x = 0, s = W = - wl + c i.e. c = W + wl and d = 0. LINEAR MEASUREMENT 31 Therefore putting constants into equations (3) and (4) gives T = -wx + W + wl T = W + w(l-x). (i.4i) and EAs = - 1 wx 2 + Wx + wlx If x = /, then and if W = 0, = Wx + ±w(2lx- x 2 ) = ml Wx + i w & lx -* 2 )] d-42) S= A[^ + I w/2 ] (1-45) S ~ 2E4 (1.44) Example 1.17 Calculate the elongation at (1) 1000 ft and (2) 3000 ft of a 3000 ft mine-shaft measuring tape hanging vertically due to its own weight. The modulus of elasticity is 30 x 10 6 lbf/in 2 ; the weight of the tape is 0*05 lbf/ft and the cross-sectional area of the tape is 0*015 in 2 . From Eq.(1.42) s = ljz[ w *+ i"(2lx-x z )] As W = 0, s= m [2lx - x * ] when x = 1000 ft / = 3000 ft 0-05 S = 2 x 30 x 10* x 0-015 C2 X 300 ° X 100 ° " 100(f] ~ °- 05 ^xl0« 2 x 30 x 10° x 0-015 when x = / = 3000 ft. ■2 From Eq. (1.44) s = — ^L 2EA 0-05x3000 2 „„„„, = 0-500 ft 2 x 30 x 10 6 x 0-015 32 SURVEYING PROBLEMS AND SOLUTIONS Example 1.18 If the same tape is standardised as 3000 ft at 451bf tension what is the true length of the shaft recorded at 2998*632 ft? I„Eq.(1.44) s- -S? -i™: ' 2EA EA i.e. T = \W where W = total weight of tape = 3000 x 0-05 = 150 Ibf Applying the tension correction, Eq.(1.39), L(J m - T s ) C = EA 3000(75-45) = 30 x 10 2 x 30 = . 200ft 30 x 10 6 x 0-015 30 x 10 6 x 0-015 .'. true length = 2998-632 + 0-2 = 2998-832 ft 1.55 Correction for sag The measuring band may be standardised in two ways, (a) on the flat or (b) in catenary. If the band is used in a manner contrary to the standard conditions some correction is necessary. (1) // standardised on the flat and used in catenary the general equation for correction is applied, viz. w 2 / 3 c = - r-^ d-45) (1.46) 24 T 2 W 2 l 24 T where w = weight of tape or wire per unit length W = wl = total weight of tape in use, T = applied tension. N.B. the units w and T must be compatible (lbf, kgf or N) Measuring head Catenary curve Fig. 1.26 Measurement in catenary (2) // standardised in catenary (a) The length of the chord may be given relative to the length LINEAR MEASUREMENT 33 of the tape or (b) the length of the tape in catenary may be given. (i) If the tape is used on the flat a positive sag correction must be applied (ii) If the tape is used in catenary at a tension T m which is different from the standard tension T s , the correction will be the difference between the two relative corrections, i.e. W z l r 1 IT C --24[n;-Tl\ (1-47) If T m > Ts the correction will be positive. (iii) If standardised in catenary using a length l a and then applied in the field at a different length l m , the correction to be applied is given as c = ?i L. //, 3 w 2 HW' h \24I 2 24T Alternatively, the equivalent tape length on the flat may be com- puted for each length and the subsequent catenary correction applied for the new supported condition, i.e. if /, is the standard length in catenary, the equivalent length on the ground = /. + c„ where c = the catenary correction. s If l m is the applied field length, then its equivalent length on the flat = W + c s ) Applying the catenary correction to this length gives lm + C = £ (l s + C j _ Cm = L +^ Thus the required correction '■m^s i — Cm I /'> 2 \ l 3 w 2 l s \24T Z I 241' j-^yi ( Z * ~ l D as Eq. (1 .48) above . 34 SURVEYING PROBLEMS AND SOLUTIONS The sag correction is an acceptable approximation based on the assumption that the measuring heads are at the same level. If the heads are at considerably different levels, Fig. 1.27, the correction should be c = c, cos 2 (l±^sin*) (1.49) the sign depending on whether the tension is applied at the upper or lower end of the tape. Measuring head Fig. 1.27 For general purposes c = c, cos 2 w 2 Pcos 2 g 24 T 2 (1.50) The weight of the tape determined in the field The catenary sag of the tape can be used to determine the weight of of the tape, Fig. 1.28 Fig. 1.28 Weight of the tape determined in the field If y is the measured sag at the mid-point, then the weight per unit length is given as STy or the amount of sag w = y = i 2 wl z ST (151) (1.52) where w = weight/unit length, y = vertical sag at the mid-point, T = applied tension, / = length of tape between supports. LINEAR MEASUREMENT 35 Example 1.19 Calculate the horizontal length between two supports, approximately level, if the recorded length is 100.237 ft, the tape weighs 15 ozf and the applied tension is 20 lbf . From Eq.(1.46) c = - W * 1 24T 2 The value of / is assumed to be 100 for ease of computation. Then 5\ 2 jjx 100 24 x 20 2 = -0-0092 ft True length = 100-2370 - 0-0092 = 100-227 8 ft Example 1.20 A 100 ft tape standardised in catenary at 25 lbf is used in the field with a tension of 20 lbf. Calculate the sag correction if w = 0-021 lbf/ft. From Eq. (1.47) c = -[1^1 T 2 T 2 ) - 100 3 x 0-021 2 / 1 _1_ 24 (20 5 "25 1 = -0-01656 i.e. -0-016 6 ft. Example 1.21 A tape 100 ft long is suspended in catenary with a ten- sion of 30 lbf. At the mid-point the sag is measured as 0-55 ft. Cal- culate the weight per ft of the tape. From Eq. (1.51), 8Ty 8 x 30 x 0-55 T - 10000 = 0-0132 IWft. Based on S.I. units these problems become 1- 19(a) Calculate the horizontal length between two supports approx- imately level if the recorded length is 30-552 2 m; the tape weighs 0-425 kgf and the applied tension is 9-072 kgf. Converting the weight and tension into units of force, 30-5522 (0-425 x 9-806 65) 2 JV VJA Z, \VftZ.D X. 3'OUD 03 V 24 (9-072 x 9-806 65) 2 36 SURVEYING PROBLEMS AND SOLUTIONS Thus there is no significance in changing the weight W and tension T into units of force, though the unit of tension must be the newton. 30-552 2 / 0-425 \ 2 24 \ 9-072/ = -0-002 8 m (-0-009 2 ft). 1.20(a) A 30*48 m tape standardised in catenary at 111-21 N is used in the field with a tension of 88-96 N. Calculate the sag correction if w= 0-0312kgf/m. Conversion of the mass/ unit length w into a total force gives 30-48 x 0-0312 x 9-80665 = 9-326 N. Eq. (1 .47) becomes c = 24 '-- -) T 2 T z \88-96 2 111-21 2 / = -30-48 x 9-326 24 = -0-00504 m (-0-016 6 ft). 1.21(a) A tape 30-48 m long is suspended in catenary with a tension of 133-446 N. At the mid-point the sag is measured as 0-168 m. Cal- culate the weight per metre of the tape. Eq. (1.51) becomes n (/ , 8 x T x y 0-816 Ty w (kgf/m) = - = — 9-80665 / 2 I 2 0-816 x 133-446 x 0-168 30-48 2 = 0-019 6 kgf/m (0-013 2 lbf/ft) Example 1.22 A tape nominally 100 ft is standardised in catenary at lOlbf and is found to be 99-933 ft. If the weight per foot is 0-01 lbf, calculate the true length of a span recorded as 49*964 ft. Standardised length = 99-933 ft Sag correction for 100 ft Q-01 2 x 100 3 rtftyIoft . c i = — 7Ta T^T - = 0-042 ft 1 24 x 10 True length on the flat = 99-975 ft LINEAR MEASUREMENT 37 True length of sub-length on flat 49-964 = 1Q0 x 99-9747 = 49-952 Sag correction for 50 ft (c oc / 3 ) = l/8c, = -0-005 True length between supports = 49- 947 ft Alternatively, by Eq.(1.48) 50 x 0-01 2 ° = 100 x 24 x 10» (5 ° " 10 ° ') = -0-018 ft ."• true length between supports = 49-964 - 0-018 = 49-946 ft. Example 1.23 A copper transmission line, %in. diameter, is stretched between two points, 1000ft apart, at the same level, with a tension of ftton,when the temperature is 90 °F. It is necessary to define its limiting positions when the temperature varies. Making use of the cor- rections for sag, temperature and elasticity normally applied to base line measurements by tape in catenary, find the tension at a tempera- ture of 10 °F and the sag in the two cases. Young's modulus for copper 10 x 10 6 lbf/in 2 , its density 555 lb/ft 3 , and its coefficient of linear expansion 9-3 x 10~ 6 per °F. (L.U.) The length of line needed = 1000 + 81 where 81 = added length due to sag = H ' 2/3 24 r 2 w = 7Tr z p lbf/ft 3-142 x 0-25 2 x 555 144 0-757 lbf/ft s; 0-757 2 x 1000 3 ol = = 19-037 ft 24 x 1120 2 i»iw/K Total length of wire = 1019-037 ft w Z 2 amount of sag v = y &T _ 0-759 x 1019 ; 8 x 1120 87-73 ft 38 SURVEYING PROBLEMS AND SOLUTIONS when temperature falls to 10 °F, Contraction of wire = Lat = 1000 x 9-3 x 10 -6 x (90-10) = 0-758 ft new length of wire = 1019-037-0-758 = 1018-279 ft as SI oc T 2 = T T 2 2 8l ± r 2 = 1120 /19-037 V18-279 = 1120 x 1-0205 = 1142 lbf Amount of sag at 10 °F (y ool) T = y, x li = 87-73 x ii?| = 86-03 ft 1.56 Reduction to mean sea level (Fig. 1.29) If the length at mean sea level is L and h = height of line above or below mean sea level, then L - R L L = R±h R±h If L = l m T c, then Fig. 1.29 Reduction to mean sea level l„R c ~ /mT /f±T (1.53) = T R±h LINEAR MEASUREMENT 39 As h is small compared with R, c = ± ~^— R If R ^ 3960 miles, (1.54) 100ft 3960 x 5280 4-8/j xl0~ 6 per 100 ft (1.55) 1.57 Reduction of ground length to grid length The local scale factor depends on the properties of the projection. Here we will consider only the Modified Transverse Mercator pro- jection as adopted by the Ordnance Survey in the British Isles. Local scale factor (F) ■(■•=) F = F o\l + „ (1.56) where F Q = the local scale factor at the central meridian, E = the Easting in metres from the true origin, p = the radius of curvature to the meridian, v = the radius of curvature at right angles to the meridian. Assuming p~v = R, then F = F I 1 + ~w) For practical purposes, F ~ F (l + 1-23 E 2 x 10" 8 ) - 0-9996013(1 + 1-23E 2 x 10~ 8 ) N.B. E = Eastings - 400 km. (1.57) (158) (1.59) 0-9996 Sub-parallel 180km Central meridian L.S.F. | ( LS - E -=^) 0-9996 Local scale error 1-0 T 0.9996 Sub-parallel 180km Distance from CM. in km Fig. 1.30 40 SURVEYING PROBLEMS AND SOLUTIONS The local scale error as shown on the graph approximates to 2R'< Example 1.24 Calculate (a) the local scale factors for each corner of the grid square TA (i.e. grid co-ordinates of S.W. corner 54), (b) the local scale factor at the centre of the square, (c) the percentage error in each case if the mean of the square corners is used instead. w fc £ & s 8 0> O) o o> o> p o TA (54) £<km) 8 jp Fig. 1.31 See Chapter 3, page 160. (a) (i) At the S.W. corner, co-ordinates are 500 km E, i.e. 100km E of central meridian. Therefore, from Eq.(1.58), L.S.F. = 0-9996013(1 + 100 2 x 1-23 x 10" 8 ) = 0-999601 + 0-000123 = 0-999724 (ii) At S.E. corner, co-ordinates are 600 km E, i.e. 200 km E of CM. .-. L.S.F. = 0-9996013(1 + 200 2 x 1-23 x 10~ 8 ) = 0-999601 + 2 2 x 0-000123 = 0-999601 + 0-000 492 = 1*000 093 (b) At centre of square, 150 km from central meridian, L.S.F. = 0-999 601 + 1'5 2 x 0-000123 = 0-999878 (c) (i) % error at S.W. corner 0-999724 - y 2 (-999724 + 1-000 093) 0-999 724 x 100 = 0-018% LINEAR MEASUREMENT 41 (ii) % error at S.E. corner 1-000 093 - 0-999908 1-000093 (iii) % error at centre = 0-999878 - 0-999908 0-999878 100 = 0-019% x 100 = 0-003% Example 1.25 Calculate the local scale factors applicable to a place E 415 km and to coal seams there at depths of 500 ft, 1000 ft, 1500 ft and 2000 ft respectively. Local radius of the earth = 6362-758 km L.S.F. = 0-9996013(1 + E 2 x 1-23 x 10" 8 ) (Eq. 1.59) Correction of length to mean sea level ~ R - h at 500ft L = / 6362-758 m 6362-758 - (500 x 0-3048 x 10~ 3 ) 6362-758 = /, 6362-758 - 0-152 . 6362-758 „ nnnnn , L 6362^606 ' 1'MOMH at 1000 ft T , 6362-758 m 6362-758 - 0-304 at 1500 ft L = I 6362 ' 758 m 6362-758 - 0-457 , 6362-758 at 2000 ft L = / 6362-758 6362-758 - 0-610 , 6362-758 ■ '""6362048 " 1 - 000096 '' At mean sea level, Easting 415 km, 42 SURVEYING PROBLEMS AND SOLUTIONS L.S.F. = 0-9996013 [1 + (415 - 400) 2 x 1-23 x 10" 8 ] = 0-9996040 at 500 ft below, L.S.F. = 0-999604 x 1-000024 = 0-999628 at 1000 ft below, L.S.F. = 0-9996040 x 1-000048 = 0-999652 at 1500 ft below, L.S.F. = 0-9996040 x 1-000072 = 0-999676 at 2000 ft below, L.S.F. = 0-999604 x 1-000096 = 0-999700. Example 1.26 An invar reference tape was compared with standard on the flat at the National Physical Laboratory at 68 °F and 201bf tension and found to be 100-024 Oft in length. The first bay of a colliery triangulation base line was measured in catenary using the reference tape and then with the invar field tape at a temperature of 60° F and with 20 lbf tension. The means of these measurements were 99*876 3 ft and 99*912 1 ft respectively. The second bay of the base line was measured in catenary using the field tape at 56 °F and 20 lbf tension and the resulting mean mea- surement was 100*213 5 ft. Given: (a) the coefficient of expansion for invar = 3-3 x 10" 7 , (b) the weight of the tape per foot run = 0-00824 lbf, (c) the inclination of the second bay = 3° 15' 00", (d) the mean height of the second bay = 820 ft A.O.D. Assuming the radius of the earth to be 20 890 000 ft, calculate the horizontal length of the second bay reduced to Ordnance Datum. (M.Q.B./S) To find the standardised length of the field tape . Reference tape on the flat at 68 °F = 100-024 Oft. Temperature correction = 100 x 3-3 x 10" 7 x (60 - 68) = -0-000 264 i.e. -0-0003 .*. Reference tape at 60 °F = 100-024 - 0-0003 = 100-023 7 ft. LINEAR MEASUREMENT 43 Thus the standardisation correction is + 0*023 7 ft per 100 ft. -w 2 / 3 Sag correction 24T 2 -(8-24 x 10~ 3 ) 2 x 100 3 - 8-24 2 9600 24 x 20 2 = -0*0071 ft The length of 100 ft is acceptable in all cases due to the close approxi- mation. The true length of the first bay thus becomes 99-8763 - 0-0071 + 0-0237 = 99-892 9 ft. The field tape applied under the same conditions when corrected for sag gives 99-9121 - 0-0071 = 99-905 Oft. The difference represents the standardisation correction 99-9050 - 99-8929 = +0-022 lft. The corrections may now be applied to the second bay . Standardisation 100-21 c = + 0-0221 x 99-91 + 0-0221 (the proportion is not necessary because of the close proximity) Temperature, L.a.(t m - t a ) c = 100 x 3-3 x 10" 7 x (56 - 60) Sag, As before as length a 100 ft Slope, L (1 - cos 6) c = - 100-21(1- cos 3° 15') = - 100-21 x 0-00161 Sea level, - Ih/R c = - 100 x 820/20890000 0-022 1 0-0001 0-0071 0-1613 0-003 9 0-1724 0-022 1 0-1503 44 SURVEYING PROBLEMS AND SOLUTIONS Horizontal length reduced to sea level = 100-2135 - 0-1503 = 100-063 2 ft. Example 1.27 The details given below refer to the measurement of the first '100ft' bay of a base line. Determine the correct length of the bay reduced to mean sea level. With the tape hanging in catenary at a tension of 20 lbf and at a mean temperature of 55° F, the recorded length was 100-082 4 ft. The difference in height between the ends was 1-52 ft and the site was 1600ft above m.s.l. The tape had previously been standardised in catenary at a tension of 15 lbf and at a temperature of 60 °F, and the distance between zeros was 100-042 ft. R = 20890000 ft. Weight of tape/ft = 0-013 lbf. Sec- tional area of tape = 0-005 6 in 2 , E = 30 x 10 6 lbf/ in 2 . Temperature coefficient of expansion of tape = 0-00000625 per 1°F. (I.C.E.) Corrections . Standardisation Tape is 100-042 ft at 15 lbf tension and 60 °F. .-. c = 0-042 ft per 100 ft Temperature c = L . a . (t m - t s ) = 100 x 6-25 x 10" 6 x (55-60) = -0-0031 Tension _ L(T m -T 9 ) A.E 100 x (20 - 15) 0-0056 x 30 x 10 6 = +0-003 Slope c = 21 1-52 2 200 0-0116 Correction + 0-0420 0-003 0-003 1 0-0116 Sag c = difference between the corrections for field and standard tensions LINEAR MEASUREMENT 45 C = - W 2 I 24 TJ 1 W c = wl = 0-013 x 100 = 1-3 lbf 1-32 2 x 100 24 = +0-013 7 15 2 - 20 2 15 2 x 20 2 Height c = hl_ R 1600 x 100 20890000 = -0-007 7 0-013 7 + 0-0587 -0-0224 + 0-0363 0-0077 0-022 4 Measured length + 100-0824 Total correction + 0-0363 Corrected length 100-118 7 ft 1.6 The Effect of Errors in Linear Measurement If the corrections previously discussed (pp. 23-40) are not applied correctly, then obviously errors will occur. Any errors within the for- mulae produce the following effects. 1.61 Standardisation Where a tape is found to deviate from standard, the error SI can be corrected in the normal way or by altering the standard temperature as previously suggested. 1.62 Malalignment and deformation of the tape (Figs. 1.32 and 1.33) (a) Malalignment. If the end of the tape is out of line by an amount d in a length /, the error will be e.g., if d= 3 in. and /= 100ft, 21 (1.60) 46 SURVEYING PROBLEMS AND SOLUTIONS - = ^L = 0-000 3 ft, JL 200 i.e. 1 in 330000. e Fig. 1.32 Malalignment of the tape (b) Deformation in the horizontal plane. If the tape is not pulled straight and the centre of the tape is out of line by d, then d d 2d ,„ _„ N 1/_| ,. e z = TTv"" -77 v = T" < L61) Jtl "Sp &L 4X1) ' &*=. 1 =**o Fig. 1.33 Deformation of the tape e.g., if d = 3 in. and 1= 100ft, e 2 = 4 x e, = 0-00123, i.e. 1 in 80000. (c) Deformation in the vertical plane. This is the same as (b) but more difficult to detect. Any obvious change in gradient can be allowed for by grading the tape or by measuring in smaller bays between these points. N.B. In (a) and (b) alignment by eye is acceptable for all purposes except very precise work. 1.63 Reading or marking the tape Tapes graduated to 0-01 ft can be read by estimation to give a pro- bable error of ± 0-001 ft. Thus if both ends o f the tape are read simultaneously the probable error in length will be V(2 x 0001 2 ), i.e. 0-001 x y/2, i.e. ±0-001 4 ft. Professor Briggs suggests that the error in setting or marking of the end of the tape is 3 times that of estimating the reading, i.e. ±0*003 ft per observation. 1.64 Errors due to wrongly recorded temperature From the correction formula c = I. a. (t m - t^), dc = ladtn (1.62) and j. = aStn (1.63) It has been suggested from practical observation that errors in re- cording the actual temperature of the tape for ground and catenary mea- surement are ± 5°F and ± 3°F respectively. LINEAR MEASUREMENT 47 If the error is not to exceed 1/10000, then from Eq. 1.63 8c 1 •= a 8t, 1UUUU i.e. 8t = I 10000 1 10000 a If a = 6-5 x 10" 6 per deg F, then 10 6 8t m = — ~ 15.4 6'5 X 10* Thus 5° produces an error of ~ 1/30000, 3° produces an error of ^ 1/50000. 1.65 Errors due to variation from the recorded value of tension These may arise from two sources: (a) Lack of standardisation of tensioning apparatus. (b) Variation in the applied tension during application (this is sig- nificant in ground taping). From the correction formula (1.39) c = H Tm ~ T *) _ LT AE ~~AE differentiation gives g c = L8T, AE (1.64) (1.65) i.e. c " r 8c 8T m T = IE < 166 > If the error is not to exceed say 1 in 10 000, then L AE 10000 • <s T AE i-e- oi m - 1O(M)0 If A = 0-003 in 2 , E = 30 x 10 a lbf/in 2 , then - T 0-003 x 30 x 10 6 8 m = — w — '~ = 91bf - i.e., an error of 1/10000 is produced by a variation of 91bf, an error of 1/30 000 is produced by a variation of 3 Ibf . The tape cross-section is x / 2 in. wide to give A = 0*003 in 2 . If the width of the tape be reduced to Kin. then, if the other dimensions 48 SURVEYING PROBLEMS AND SOLUTIONS remain constant, the cross-sectional area is reduced to %A = 0*000 8 in 2 . In this case a variation of 3 lbf will produce an error of 1/10 000 and the accuracy will be reduced as the cross-sectional area dimini- shes. 1.66 Errors from sag Where the tape has been standardised on the flat and is then used in catenary with the measuring heads at different levels, the approxi- mation formula is given as -/ 3 w 2 cos 2 6 c = (1.50) 24I 2 where 6 is the angle of inclination of the chord between measuring heads. The value of cos 2 becomes negligible when 6 is small. The sources of error are derived from: (a) an error in the weight of the tape per unit length, w, (b) an error in the angular value, B, (c) an error in the tension applied, T. By successive differentiation, dc w = - 2/ 3 wcos 2 flg W (1 67) 24 T 2 2cdW (1.68) i.e. w 8c„, 28w c w (1.69) This may be due to an error in the measurement of the weight of the tape or due to foreign matter on the tape, e.g. dirt. -l 3 w 2 8c e = — 2 sin20S0 See c Sc T = = 2c tan 6 86 = 2tan0S0 2/ 3 w 2 cos 2 0ST 24T : 2c8T T 8c T _ -28T c " T (1.70) (1.71) (1.72) (1.73) (1.74) (1.75) LINEAR MEASUREMENT 49 The compounded effect of a variation in tension gives 2l*w 2 cos 2 (9 8T 18T 24T 2 + ~AE < L76 ) Example 1.28 If / = 100 ft, w = 0-01 ± 0-001 lbf per ft, $ = 2° ± 10" T = 10 ± 1 lbf, 100 3 x 0-01 2 x cos 2 2° cos 2 2° ~24 C = 24xl0 2 = ^lf~ = 0>04161ft - Then~ 2c 8w w = ~w~ = 2 x °* 04161 x O'l = 0-008 32 ft i.e. 10% error in weight produces an error of 1/12000. « 2c tan fl 86" 0-08322 806 = 206 265 = "2^6167 x °-° 524 x 10 = 0-000 000 21 ft. This is obviously negligible. Sc t = ~y- = 0-083 22 xO-U 0-008 322 ft i.e. 10% error in tension produces an error of 1/12000. Example 1.29 A base line is measured and subsequent calculations show that its total length is 4638-00 ft. It is later discovered that the tension was recorded incorrectly, the proper figure being 10 lbf less than that stated in the field book, extracts from which are given below. Assuming that the base line was measured in 46 bays of nominal length 100 ft and one bay of nominal length 38 ft, calculate the error incurred in ft. Extract from field notes Standardisation temperature = 50 °F Standardisation tension = 20 lbf Measured temperature = 45 op Measured tension _ 40 lbf Young's modulus of tape = 30 x 10 6 lbf/in 2 Cross-sectional area of tape = 0-125 in. x 0-05 in.- Weight of lin 3 of steel = 0-28 lbf. (N.U.) Weight of steel tape per ft = 0-125 x 0-05 x 12 x 0-28 = 0-021 lbf. 50 SURVEYING PROBLEMS AND SOLUTIONS From Eq. (1.39) c = L(Jm ' J s) Then the error due to wrongly applied tension = c - c' = L(T m - r.) _ L(T^ - T s ) AE AE = M. (T - T') AE K m m) where T m = true applied tension, T' m = assumed applied tension. Error 4638 ( 30 - 40 > . . -0-24736ft. 0-125 x 0-05 x 30 x 10 6 -W z i From Eq. (1.46) correction for sag c = — — -j .'. Error due to wrongly applied tension = c - c, w!i/i _ i_\ 24 \T Z Tfj W = 100 x 0-021 = 2-1 lbf Error for 100 ft bay 2-1 2 x 100 /Jl_ _ _1_\ 24 Uo 2 40 2 / 441 / l600 - 900 \ = ~ 24 \l600 x 900/ _ 441 / 700 \ 24 \1 440 000/ = -0-00893 Error for 46 bays = -0-410 78 ft Error for 38 ft bay W = 38 x 0-021 = 0-798 lbf _ _ 0-798 2 x 38 / 700 \ n0t ~ 24 \l 440 000/ = - 0-000 49 ft .-. Total error for sag = -0-411 27 ft Total error for tension = - 0-247 36 ft LINEAR MEASUREMENT 51 Total error = - 0-658 63 ft i.e. Apparent reduced length is 0*658 6 ft too large. 1.67 Inaccurate reduction to the horizontal The inclined length may be reduced by obtaining (a) the difference in level of the measuring heads or (b) the angle of inclination of the tape, (a) The approximation formula is given as (Eq. 1.15) d 2 C =2l ...) Adopting the first term only, from the differentiation 8c _ d8d I 2c 8d d 8c c 28d d (1.77) (1.78) (1.79) If 8d/d= 1%, when / = 100 and d= 5 ± 0-05 ft, * 2 x 5 2 x 0-01 8c = o T7^ = ± 0-002 5 f t i.e. 1 in 40000. 2 x 100 As the difference in level can be obtained without difficulty to ± 0-01 ft, 8c = ± 0-000 5 ft, i .e . 1/200 000 . (b) By trigonometrical observations (Eq. 1.12), c = 1(1 - cos0) Then 8c = /sin0S0 (1.80) _ /sinflgfl" " 206265 C1,81) i.e. 8c oc sin0 86 (1.82) If / = 100, B = 30° ± 20", 8c = ± 10 ° * °; 5 * 2 ° - ±0-0048ft i.e. 1/20000. 2Uo 265 The accuracy obviously improves as 6 is reduced. As the angle of inclination increases the accuracy in the measure- ment of 6 must improve. 52 SURVEYING PROBLEMS AND SOLUTIONS 1.68 Errors in reduction from height above or below mean sea level From the formula Ih by differentiating 8c = (1.83) R. cSh = T£- (1.84) The % error in the correction is equal to the % error in the height above or below M.S.L. 1.69 Errors due to the difference between ground and grid distances Local scale factor is given by Eq.(1.59) 0-9996013(1 + 1-23 E 2 x 10~ 8 ) where E is the distance in km from the central meridian (i.e. the East- ings - 400 km). As this amounts to a maximum of 0*04% it is only effective in pre- cise surveys . Exercises 1(b) 10. A 300ft tape has been standardised at 80 °F and its true length at this temperature is 300*023 ft. A line is measured at 75 °F and re- corded as 3486*940 ft. Find its true length assuming the coefficient of linear expansion is 6*2 x 10 -6 per deg F. (Ans. 3487- 10 ft) 11. A base line is found to be 10560 ft long when measured in caten- ary using a tape 300ft long which is standard without tension at 60 °F. The tape in cross-section is 1/8 x 1/20 in. If one half of the line is measured at 70 °F and the other half at 80 °F with an applied tension of 501bf, and the bays are approximately equal, find the total correction to be applied to the measured length. Coefficient of linear expansion = 6*5 x 10 -6 per deg F. Weight of 1 in 3 of steel = 0-28 lbf . Young's modulus = 29 x 10 6 lbf/in 2 . (Ans. -3 -042 ft) 12. A 100 ft steel tape without tension is of standard length when placed on the ground horizontally at a temperature of 60 °F. The cross- sectional area is 0*0103 in 2 and its weight 3*49 lbf, with a coefficient of linear expansion of 6*5 x 10~ 6 per deg F. The tape is used in the field in catenary with a middle support such LINEAR MEASUREMENT 53 that all the supports are at the same level. Calculate the actual length between the measuring heads if the temperature is 75 °F and the tension is 20 lbf. (Assume Young's mod- ulus 30 x 10 6 lbf/in 2 ). (Ans. 99-985 lft) 13. A nominal distance of 100 ft was set out with a 100 ft steel tape from a mark on the top of one peg to a mark on the top of another, the tape being in catenary under a pull of 20 lbf and at a mean temperature of 70 °F. The top of one peg was 0*56 ft below the top of the other. The tape had been standardised in catenary under a pull of 25 lbf at a temperature of 62 °F. Calculate the exact horizontal distance between the marks on the two pegs and reduce it to mean sea level. The top of the higher peg was 800ft above mean sea level. (Radius of earth = 20:9 x 10 6 ft ; density of tape 0*28 lbf/in 3 ; section of tape = 0*125 x 0*05 in.; Young's modulus 30 x 10 6 lbf/in 2 ; coefficient of expansion 6' 25 x 10" 6 per 1° F) (I.C.E. Ans. 99-980 4 ft) 14. A steel tape is found to be 299-956 ft long at 58 °F under a ten- sion of 12 lbf. The tape has the following specifications: Width 0-4 in. Thickness 0-018 in. Young's modulus of elasticity 30 x 10 5 lbf/in 2 Coefficient of thermal expansion 6-25 x 10" 6 per deg F. Determine the tension to be applied to the tape to give a length of precisely 300 ft at a temperature of 68 °F. (N.U. Ans. 30 lbf) 15. (a) Calculate to three decimal places the sag correction for a 300 ft tape used in catenary in three equal spans if the tape weighs 1 lb/100 ft and it is used under a tension of 20 lbf. (b) It is desired to find the weight of a tape by measuring its sag when suspended in catenary with both ends level. If the tape is 100 ft long and the sag amounts to 9-375 in. at mid-span under a tension of 20 lbf, what is its weight in ozf per 100 ft? (N.U. Ans. 0-031 ft, 20 ozf) 16. Describe the methods used for the measurement of the depth of vertical mine shafts and discuss the possible application of electronic distance measuring equipment. Calculate the elongation of a shaft measuring tape due to its own weight at (1) 1000 ft and (2) 3000 ft, given that the modulus of elastici- ty is 30 x 10 6 lbf/in 2 } weight of the tape 0-051bf/ft run, and the cross- sectional area of the tape 0-015 in 2 . (N.U. Ans. 0-055 6 ft, 0-500 ft) 54 SURVEYING PROBLEMS AND SOLUTIONS 17. (a) Describe the measuring and straining tripods used in geodelic base measurement. (b) The difference between the readings on a steel tape at the terminals of a bay between which it is freely suspended was 94*007 ft, the tension applied being 201bf, the temperature 39*5 °F, and the height difference between the terminals 5*87 ft. The bay was 630 ft above mean sea level. If the tape, standardised on the flat, measured correctly at 68 °F under lOlbf tension, and its weight was 0*0175 lbf per ft, its coefficient of expansion 0*62 x 10~ 8 per deg F and its coefficient of extension 0*67 x 10" 5 per lb, calculate the length of the bay reduced to mean sea level. (Radius of earth = 20-9 x 10 6 ft). (L.U. Ans. 93-784 ft) 18. The steel band of nominal length 100ft used in the catenary mea- surement of a colliery base line, has the following specification: (i) Length 100*025 ft at 10 lbf tension and 68 °F. (ii) Sectional area 0*004 in 2 (iii) Weight 22 ozf . (iv) Coefficient of linear expansion 6*25 x 10~ 6 per deg F. (v) Modulus of elasticity 30 x 10* lbf/in 2 . The base line was measured in 10 bays and the undernoted obser- vations were recorded in respect of the first five which were of average height 625 ft above Ordnance Datum. Observed Bay Bay Level Tension Bay Length Temperature Difference Applied 1 100*005 52 °F 0*64 20 lbf 2 99*983 54°F 1*23 20 lbf 3 100*067 54 °F 0*01 20 lbf 4 100*018 58 °F 0*79 20 lbf 5 99*992 60°F 2*14 20 lbf Correct the bays for standard, temperature, tension, sag, slope, and height above Ordnance Datum and compute the corrected length of this part of the base line. Take the mean radius of the earth to be 20890000 ft. (M.Q.B./S Ans. 500'044ft) 19. The steel band used in the catenary measurement of the base line of a colliery triangulation survey has the undernoted specification: (i) length 50*000 3 m at a tension of 25 lbf at 60° F. (ii) weight 2*5 lbf (iii) coefficient of linear expansion 6*25 x 10~ 8 per deg F. LINEAR MEASUREMENT 55 The undernoted data apply to the measurement of one bay of the base line: (i) length 50-002 7 m (ii) mean temperature 53 °F (iii) tension applied 25 lbf (iv) difference in level between ends of bay 0*834 m (v) mean height of bay above mean sea level 255*4 m. Correct the measured length of the bay for standard, temperature, sag, slope, and height above mean sea level. Assume the mean radius of the earth is 6-37 x 10 6 m. (M.Q.B./S Ans. 49-971 m) 20. A base line was measured with an invar tape 100 ft long which had been standardised on the flat under a tensile load of 15 lbf and at a temperature of 60 °F. Prior to the measurement of the base line the tape was tested under these conditions and found to record 0*015 ft too much on the standard length of 100 ft. The base line was then divided into bays and the results obtained from the measurement of the bays with the tape suspended are shown below: Bay Length (ft) Difference in level between supports (ft) Air temperature (°F) 1 99*768 2*15 49*6 2 99*912 1*62 49*6 3 100*018 3*90 49*8 4 100*260 4*28 50*2 5 65*715 0*90 50*3 Modulus of elasticity (E) for invar = 22 x 10 6 lbf/in 2 . Coefficient of linear expansion of invar = 5*2 x 10~ 7 per deg F. Field pull = 25 lbf. Cross-sectional area of tape = 0*004 in 2 . Weight per ft run of tape = 0*010 2 lbf. Average reduced level of base line site = 754*5 ft. Radius of earth = 20*8 x 10 6 ft. Correct the above readings and determine to the nearest 0*001 ft the length of the base line at mean sea level. (I.C.E. Ans. 465*397 ft) 21. The following readings were taken in measuring a base line with a steel tape suspended in catenary in five spans: 56 SURVEYING PROBLEMS AND SOLUTIONS pan Mean reading Difference in level Tension Mean of tape (ft) between Index Marks (ft) (Ibf) temperature (°F) 1 100.155 3-1 25 73 2 100*140 0*9 50 76 3 100«060 1-2 25 78 4 100» 108 3-1 25 80 5 100-182 2-0 25 80 The tape reading was 100*005 ft when calibrated in catenary under a tension of 25 lbf at a temperature of 65 °F between two points at the same level precisely 100 ft apart. Other tape constants are: width of tape = 0*250 in; thickness of tape = 0*010 in; weight of steel = 0*283 lbf/in 3 ; E for steel = 30 x 10 6 lbf/in 2 ; coefficient of expansion of steel = 6*2 x 10" per deg F. Compute the length of the base line. (I.C.E. Ans. 500*568 ft) 22. A short base line is measured in four bays with a 100ft invar band in catenary under a pull of 20 lbf with the following field readings: Bay 1 2 3 4 t°F 65*2 64*0 65*5 63*8 hit 5*08 1*31 2*31 2*13 I ft 99*6480 99*751 7 99*541 7 99*9377 where t is the field temperature, h is the difference in level between the ends of each bay and / is the mean reading of the invar band . When standardised in catenary under a pull of 20 lbf at 68*5°F the standard length of the invar band was 99*999 ft and the mean altitude of the base is 221 ft above sea level. If the coefficient of expansion of invar is 0*000 0003 per deg F and the radius of the earth is 20*9 x 10* ft what is the length of the base line reduced to sea level? (I.C.E. Ans. 398*6829 ft) Bibliography CLARK, D., Plane and Geodetic Surveying, Vols. I and II (Constable) HOLLAND, J.L., WARDELL, K. and WEBSTER, A.G., Surveying, Vols. I and II. Coal Mining Series (Virtue) SMIRNOFF, M.V., Measurement for Engineering and Other Surveys (Prentice-Hall) BANNISTER, A. and RAYMOND, S., Surveying (Pitman) THOMAS, W.N., Surveying, 5th ed. (Edward Arnold) BRIGGS, N., The Effects of Errors in Surveying (Griffin) MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.) ORDNANCE SURVEY publication, Constants, Formulae and Methods used in Transverse Mercator (H.M.S.O.) 2 SURVEYING TRIGONOMETRY 'Who conquers the triangle half conquers his subject' M.H. Haddock Of all the branches of mathematics, trigonometry is the most im- portant to the surveyor, forming the essential basis of all calculations and computation processes. It is therefore essential that a thorough working knowledge is acquired and this chapter is an attempt to sum- marize the basic requirements. 2.1 Angular Measurement There are two ways of dividing the circle: (a) the degree system, (b) the continental 'grade' system. The latter divides the circle into 4 quadrants of 100 grades each and thereafter subdivides on a decimal system. It has little to com- mend it apart from its decimalisation which could be applied equally to the degree system. It has found little favour and will not be con- sidered here. 2.11 The degree system 180 Clockwise rotation used by surveyors Anticlockwise rotation used by mathematicians Fig. 2.1 Comparison of notations 57 58 SURVEYING PROBLEMS AND SOLUTIONS The circle is divided into 360 equal parts or degrees, each degree into 60 minutes, and each minute into 60 seconds. The following sym- bols are used:— degrees (°) minutes ( ') seconds (") so that 47 degrees 26 minutes 6 seconds is written as 47° 26' 06" N.B. The use of 06" is preferred in surveying so as to remove any doubts in recorded or computed values. In mathematics the angle is assumed to rotate anti-clockwise whilst in surveying the direction of rotation is assumed clockwise. This variance in no way alters the subsequent calculations but is merely a different notation. 2.12 Trigonometrical ratios (Fig. 2.2) Assume radius = 1 Cotangent© a) C Versine Fig. 2.2 Sine (abbreviated sin) angle = 4| = 4p = AB = GO Cosine (abbreviated cos) angle = 9J* = 93. = OB = GA OA 1 Tangent (abbreviated tan) angle = M. = sin_0 = DC = DC OB cos OC 1 Cotangent 6 (cot 0) = _ cos 6_ OB _ FE FE tan 6 sin 6 AB FO SURVEYING TRIGONOMETRY Cosecant (cosec 0) = —. — ■* = -t-~ = sin0 Ad OE OF OE 1 Secant (sec 0) - * - °A - cos OB OD OC OD 1 59 Versine (vers 0) = 1 - cos = OC - OB Coversine (covers 0) = 1 - sin0 = OF - OG. N.B. In mathematical shorthand sin -1 x means the angle (x) whose sine is ... . If OA = radius = 1 then, by Pythagoras, sin 2 + cos 2 = 1. (2.1) sin 2 = 1 - cos 2 (2.2) cos 2 = 1 - sin 2 (2.3) Dividing Eq. 2.1 by cos 2 0, sin 2 cos 2 1 + cos 2 cos 2 cos 2 i.e. tan 2 + 1 = sec 2 tan 2 = sec 2 - 1. (2.4) Dividing Eq. 2.1 by sin 2 0, sin 2 cos 2 1 sin 2 sin 2 sin 2 i.e. 1 + cot 2 = cosec 2 (2.5) sin0 = V(l- c os 2 0). (2.6) cos0 = V(l-sin 2 0). (2.7) (2.8) (2.9) (2.10) (2.11) tan0 = sin0 sin0 cos0 V(l-sin 2 0) V(l - cos 2 0) or = - 5Li - COS0 cos — 1 1 sec0 VCi + tan 2 ^) sin0 — 1 1 cosec \/(l + cot 2 0) which shows that, by manipulating the equations, any function can be expressed in terms of any other function. 60 SURVEYING PROBLEMS AND SOLUTIONS 2.13 Complementary angles The complement of an acute angle is the difference between the angle and 90°, i.e. if angle A 30 ( its complement = 90° - 30° = 60° The sine of an angle The cosine of an angle The tangent of an angle The secant of an angle The cosecant of an angle 2.14 Supplementary angles cosine of its complement sine of its complement cotangent of its complement cosecant of its complement secant of its complement The supplement of an angle is the difference between the angle and 180°, i.e. if angle A = 30° its supplement = 180° - 30°= 150°. The sine of an angle = sine of its supplement cosine of an angle = cosine of its supplement (but a nega- tive value) tangent of an angle = tangent of its supplement (but a nega- tive value) These relationships are best illustrated by graphs. Sine Graph (Fig. 2.3) 180 Fig. 2. 3 The sine graph Let the line OA of length 1 rotate anticlockwise. Then the height above the horizontal axis represents the sine of the angle of rotation. At 90° it reaches a maximum = 1 At 180° it returns to the axis. At 270° it reaches a minimum = -1 It can be seen from the graph that SURVEYING TRIGONOMETRY 61 sin 30° = sin (180 - 30), i.e. sin 150° = -sin (180 + 30), i.e. -sin 210° = -sin (360 -30), i.e. -sin 330° Thus the sine of all angles - 180° are +ve (positive) the sine of all angles 180 - 360 are -ve (negative). Cosine Graph (Fig. 2.4). This is the same as the sine graph but dis- placed by 90°. Fig. 2. 4 The cosine graph cos 30° = -cos (180 -30) = -cos 150° = -cos (180 + 30) = -cos 210° = +cos(360-30) = + cos 330° Thus the cosine of all angles - 90° and 270° -360° are + ve 90° -270° ate-ve Tangent Graph (Fig. 2.5) This is discontinuous as shown. ■too +oo -hoo -oo -oo Fig. 2.5 The tangent graph 62 SURVEYING PROBLEMS AND SOLUTIONS tan 30° = tan (180 +30), i.e. tan 210° = -tan (180 -30), i.e. -tan 150° = -tan (360 -30), i.e. -tan 330° Thus the tangents of all angles 0-90° and 180°- 270° ate + ve 90°- 180° and 270°- 360° are -ve Comparing these values based on the clockwise notation the sign of the function can be seen from Fig. 2.6. i*£ 1st quodrtmt 0*-90* .+ 2nd quadrant 90-180 • f%ir\* 3rd quadrant 160-270 • 4*A*«* * 4th quadrant 270-360 Fig. 2.6 Let the rotating arm be +ve 6° 1st quadrant sin 0, SURVEYING TRIGONOMETRY sin COS0, tan0, —, i.e. + ±, i + 2nd quadrant 2 = (180-0) 3rd quadrant 3 = (0-180) 4th quadrant 4 = (360-0) sin0 2 *» ± + cos 2 = jr. + tan 2 = — sin 3 s= jz + COS0 3 ;~ tan 3 = sin 4 = cos0 4 = tan 4 = .e. + .e. + i.e. + i.e. - i*e. - i.e. ■-* i.e. - i.e. + i.e. - — , i.e. + i.e. 63 2.15 Basis of tables of trigonometrical functions Trigonometrical tables may be prepared, based on the following series : ~ /}3 /J5 /j7 " H R R (2.12) sin0 = 0-£! + i!-£ + 3! 5! 7! 3! 5! 7! where is expressed as radians, see p. 72. and 3! is factorial 3, i.e. 3 x 2 x 1 5! is factorial 5, i.e. 5x4x3x2x1. cos0 = 1 _ jf? + §1 _ §1 + _.. (2.13) 2! 4! 6! This information is readily available in many varied forms and 64 SURVEYING PROBLEMS AND SOLUTIONS to the number of places of decimals required for the particular prob- lem in hand. The following number of places of decimals are recommended:— for degrees only, 4 places of decimals, for degrees and minutes, 5 places of decimals, for degrees, minutes and seconds, 6 places of decimals. 2.16 Trigonometric ratios of common angles The following basic angles may be calculated. B ' 30° 30° \ 2/ \ 2 >/3 \ /\ 60 ° ^ 6oy \ A 1 D 1 C Fig. 2.7 Trigonometrical ratios of 30° and 60 From the figure, with BD perpendicular to AC, Let AB = BC = AC = 2 units, then AD = DC = 1 unit, by Pythagoras BD = y/(2 2 -l z ) = y/3. Thus sin 30° =1=0-5 = cos 60° 2 sin 60° = tan 30° = V3 1-732 0-8660 cos 30° T ~ 2 _L = V? 1-732 V3 3 3 = 0-577 3 = cot 60° tan 60° = ^ = 1-7320 = cot 30° Similarly values for 45° may be obtained. Using a right-angled isosceles triangle where AC = BC = 1, by Pythagoras AB = yj{\ 2 + l 2 ) = V 2 SURVEYING TRIGONOMETRY 65 Fig. 2.8 Trigonometrical ratios of 45° 1 = )il 1-414 2 0-7071 = cos 45° cot 45° It can now be seen from the above that sin 120° = sin (180 -120) = sin 60° = 0-8660 /hereas cos 120° = -cos (180 - 120) = -cos 60° = -0-5 tanl20° = -tan (180 - 120) = -tan60° = -1*7320 vlso sin 210° = -sin (210 -180) = -sin 30° = -0-5 cos240° = - cos (240 - 180) = -cos60° = -0-5 tan 225° = +tan (225 - 180) = tan 45° = 1-0 sin330° = - sin (360 - 330) = -sin30° = -0-5 cos 315° = +cos(360-315) = + cos 45° = 0-7071 tan 300° = - tan (360 - 300) = -tan 60° = -1-7320 17 Points of the compass (Fig. 2.9) These are not used in Surveying but are replaced by Quadrant (or uadrantal) Bearings where the prefix is always N or S with the suffix or W, Fig. 2.10. g. NNE = N 22° 30' E. ENE = N 67° 30' E. ESE = S 67° 30' E. SSE = S 22° 30' E. SW = S 45° 00' W. NW = N 45° 00' W. 66 SURVEYING PROBLEMS AND SOLUTIONS Fig. 2.9 Points of the compass NW»N45°W SW = S45° W NNE = N22° 30' E ENE = N67° 30'E ESE =S67° 30'E SSE =S22° 30'E SURVEYING TRIGONOMETRY 67 2.18 Easy problems based on the solution of the right-angled triangle N.B. An angle of elevation is an angle measured in the vertical plane where the object is above eye level, i.e. a positive vertical angle, Fig. 2.11. An angle of depression is an angle measured in the vertical plane where the object is below eye level, i.e. a negative vertical angle. Fig. 2. 11 Vertical angles In any triangle there are six parts, 3 sides and 3 angles The usual notation is to let the side opposite the angle A be a etc, as shown in Fig. 2.12. The following facts are thus known about the given right-angled triangle ABC. Angle C = 90° Angle A + Angle fi = 90° c 2 = a 2 + b z (by Pythagoras) sin A = SL c cos A = k c tan A = £ b Fig. 2.12 68 SURVEYING PROBLEMS AND SOLUTIONS tan A = a c - x -r c b i.e. sin A ~- cos A. To find the remaining parts of the triangle it is necessary to know 3 parts (in the case of the right-angled triangle, one angle = 90° and therefore only 2 other facts are required). Example 2.1. In a right-angled triangle ABC, the hypotenuse AB is 10 metres long, whilst angle A is 70°. Calculate the remaining parts of the triangle. As the hypotenuse is AB (c) the right angle is at C (Fig. 2.13). then a c = sin 70° a = c sin 70° = 10 sin 70° = 10 x 0-939 69 = 9-397 metres b c = cos 70° b = c cos 70° = 10 x 0-34202 = 3-420 metres Check Fig. 2. 13 AngleS = 90° - 70° = 20° A = tanS = l'*®l = 0-363 97 a 9-396 9 i.e. angle B = 20° 00' Example 2.2 It is necessary to climb a vertical wall 45 ft (13-7 m) high with a ladder 50 ft (15-2 m) long, Fig. 2.14. Find (a) How far from the foot of the wall the ladder must be placed, (b) the inclination of the ladder || = sin^ = 0-9 50 angle A = 64° 09' 30" thus angle B = 25° 50' 30" £ = cos A thus b = 50 cos 64° 09' 30" = 21-79 ft (6-63 m) Ans. (a) 21-79 ft (b) 64° 09' 30" from the horizontal. SURVEYING TRIGONOMETRY B 45' 69 Fig. 2.14 Example 2.3. A ship sails 30 miles (48-28 km) on a bearing N 30° E. It then changes course and sails a further 50 miles (80'4 km) N 45° W. Find (a) the bearing back to its starting point, (b) the distance back to its starting point. N.B. See chapter 3 on bearings To solve this problem two triangles, ADB and BCE, are joined to form a resultant third ACF (Fig. 2.15). In triangle ADB, AB is N30°E 30 miles (48-28 km). The distance travelled N = AD. but 4§ = cos 30° AB then AD = 30 cos 30° = 25-98 miles (41 -812 km) The distance travelled E = DB but M = sin 30° AB .-. DB = 30 sin 30° = 15-00 miles (24 -140 km) Similarly in triangle BCE the dis- tance travelled N = BE but BE = BC cos 45° = 50 cos 45° = 35-35 miles (56-890 km) The distance travelled W (CE) = The distance travelled N = 35-35 miles as the bearing = 45° (sin 45° = cos 45°). In resultant triangle ACF, 70 SURVEYING PROBLEMS AND SOLUTIONS CF = CE - DB = 35-35 - 15-00 = 20-35 miles (32-750) AF = AD + BE = 25-98 + 35-35 = 61-33 miles (98-702) tantf = CF = JP-^I = 0-33181 AF 61-33 6 = 18° 21 ' 20" .-. bearing A C = N 18° 21' 20" W. AC AF cos 6 61-33 cos 18° 21' 20 '-, — Tl = 64-62 miles (104-0 km) Example 2.4. An angle of elevation of 45° was observed to the top of a tower. 42 metres nearer to the tower a further angle of elevation of 60° was observed. Find (a) the height of the tower, (b) the distance the observer is from the foot of the tower. Fig. 2. 16 InF] ig.2.16, AC H = cot A i.e. AC = H cot A also BC H = cotS i.e. BC = H cot B. AC - BC AB = ft (cot ,4 - cotB) H AB cot A - cot B SURVEYING TRIGONOMETRY 71 42 cot 45° -cot 60° 42 1-0-577 4 — 0-4226 - 99 " 38ln BC = H cot B = 99-38 cot 60° = 99-38 x 0-577 4 = 57-38 m AC = DC = 99-38 BC = AC - AB = 99-38 - 42 = 57-38 m Check Exercises 2(a) 1. A flagstaff 90 ft high is held up by ropes, each being attached to the top of the flagstaff and to a peg in the ground and inclined at 30° to the vertical; find the lengths of the ropes and the distances of the pegs from the foot of the flagstaff. (Ans. 103-92 ft, 51-96 ft) 2. From the top of a mast of a ship 75 ft high the angle of depression of an object is 20°. Find the distance of the object from the ship. (Ans. 206 -06ft) 3. A tower has an elevation 60° from a point due north of it and 45° from a point due south. If the two points are 200 metres apart, find the height of the tower and its distance from each point of observation. (Ans. 126-8 m, 73-2 m, 126-8 m) 4. A boat is 1500 ft from the foot of a vertical cliff. To the top of the cliff and the top of a building standing on the edge of the cliff, angles of elevation were observed as 30° and 33° respectively. Find the height of the building to the nearest foot. (Ans. 108 ft) 5. A vertical stick 3 m long casts a shadow from the sun of 1-75 m. What is the elevation of the sun ? (Ans. 59° 45') 6. X and Y start walking in directions N17°W and N73°E; find their distance apart after three hours and the direction of the line joining them. X walks at 3 km an hour and Y at 4 km an hour. (Ans. 15 km S70°08'E) 7. A, B, and C are three places. B is 30 km N67H°E of A, and C 72 SURVEYING PROBLEMS AND SOLUTIONS is 40 km S22 1 / 2 °E of B. Find the distance and bearing of C from A. (Ans. 50 km, S59°22'E) 2.2 Circular Measure The circumference of a circle = 2rrr where 77 = 3*1416 approx. 2.21 The radian The angle subtended at the centre of a circle by an arc equal in length to the radius is known as a radian. Thus 2 77 radians = 360° /. 1 radian = M 277 = 57°17 , 45" approx. = 206 265 seconds. This last constant factor is of vital importance to small angle cal- culations and for conversion of degrees to radians. Example 2.5. Convert 64° 11' 33" to radians. 64° 11' 33" = 231093 seconds. .-. no. of radians = 231093 = 1-120 37 rad. 206 265 Tables of radian measure are available for 0°-90° and, as the radian measure is directly proportional to the angle, any combination of values produces the same answer for any angular amount. By tables, 64° = 1-11701 11' = 0-003 20 33" = 0-000 16 64° 11' 33" — 1-120 37 It now follows that the length of an arc of a circle of radius r and subtending d radians at the centre of the circle can be written as arc = r.6 rad (2.14) This is generally superior to the use of the formula ■* - 2m * m (21S) N.B. When 6 is written it implies 6 radians. To find the area of a circle. A regular polygon ABC ... A is drawn inside a circle, Fig. 2.17. SURVEYING TRIGONOMETRY 73 Draw OX perpendicular to AB Then area of polygon = ±OX(AB + BC + ...) = ■tOX (perimeter of polygon) When the number of sides of the polygon is increased to infinity (oo), OX becomes the radius, the perimeter becomes the cir- cumference, and the polygon becomes the circle .*. area of circle = V2.r .2ttt = TFT 2 The area of the sector OAB. area of sector e area of circle 2tt ari=>a of sftrtnr ^ rrr " _ 2tt \r*Q Fig. 2. 17 (2.16) 2.22 Small angles and approximations For any angle < 90° (i.e. < 77/2 radians) tantf > > sim A Fig. 2. 18 Let angle AOC = 6 OA = OC = r and let AB be a tangent to the arc AC at A to cut OC produced at B. Draw AD perpendicular (1) to OB. 74 SURVEYING PROBLEMS AND SOLUTIONS Then area of triangle OAB = ±0A . AB = h.rtand = ±r 2 tan0 2 2 area of sector OAC = |r 2 area of triangle OAC = ±OC.AD = Ir.r sin0 = |r 2 sin0 Now triangle OAB > sector (X4C > triangle OAC. Ar 2 tan0 > |r 2 > |r 2 sin0 tan0 > 9 > sin0 This is obviously true for all values of < it/2. Take to be very small. Divide each term by sin0, then * > , 3 > 1 cos0 sin0 It is known that as -* then cos0 -* 1. Thus cos0 ~ 1 when is small .*. .. must also be nearly 1. sin0 The result shows that sin0 may be replaced by 0. Similarly, dividing each term by tan0, 1 > A > s * n fl tan0 tan0 i.e. 1 > " > cos0 tan0 Tan0 may also be replaced by 6. It can thus be shown that for very small angles tan0 =2.. ^ sin0. The following values are taken from H.M. Nautical Almanac Office. Five-figure Tables of Natural Trigonometrical Functions. (These tables are very suitable for most machine calculations.) SURVEYING TRIGONOMETRY 75 Angle Tangent Radian Sine 1°00'00" 0-01746 0-01745 0-01745 1° 30' 00" 0-026 19 0-02618 0-026 18 2° 00' 00" 0-03492 0-03491 0-03490 2° 30' 00" 0-04366 0-04363 0-04362 3° 00' 00" 0-05241 0-05236 0-05234 3° 30' 00" 0-061 16 0-06109 0-06105 4° 00' 00" 0-06993 0-06981 0-06976 4° 30' 00" 0-07870 0-078 54 0-078 46 5° 00 '00" 0-08749 0-08727 0-08716 From these it can be seen that $ may be substituted for sin 6 or tan0 to 5 figures up to 2°, whilst 6 may be substituted for sin0 up to 5° and for tantf up to 4° to 4 figures, thus allowing approxima- tions to be made when angles are less than 4° Example 2.6. If the distance from the earth to the moon be 250000 miles (402000 km) and the angle subtended 0° 30', find the diameter of the moon. 250 OOO miles Fig. 2. 19 The diameter ~ arc ABC 2L 250000 x 30'rad - 250000 v 30 x 60 206265 =2 2180 miles (3510 km) Example 2.7. Find as exactly as possible from Chambers Mathemati- cal Tables the logarithmic sines of the following angles: A = 00° 02' 42" and C = 00° 11' 30" Use these values to find the lengths of the sides AB and AC in a tri- angle ABC when BC = 12-736 ft. Thereafter check your answer by another method, avoiding as far as possible using the tables at the same places as in the first method. The lengths are to be stated to three places of decimals. (M.Q.B/S) 76 SURVEYING PROBLEMS AND SOLUTIONS As the sines and tangents of small angles change so rapidly, special methods are necessary. Method 1 Chambers Mathematical Tables give the following method of finding the logarithmic sine of a small arc: To the logarithm of the arc reduced to seconds, add 4-685 574 9 and from the sum subtract 1/3 of its logarithmic secant, the index of the latter logarithm being previously diminished by 10. 00°02'42" = 162" logl62 2-2095150 constant 4-6855749 -|(logsec02'42" - 10) = | x 0-000 000 2 00° 11" 30" = 690" -!(log sec 11' 30" - 10) = I x 0-0000024 Method 2 As the sines and tangents of small angles approximate to the value, radian value of A 00° 02' 42" = * 6 ^ = 0-000785 3974 206 265 logO-000785397 4 = 6-8950895 radian value of B 00° 11' 30" = „ 690 „ = 0-003 345 2112 206 265 logO-003 3452112 = 7-524423 5 (2) Vegas tables (to l") 6-895089 5 6-8950898 7-524 4235 7-524 4231 To find the length of sides AB and AC when BC = 12-736 : AB = BC sin C = 12-736 sin (02' 42" + 11' 30") (see 2.51 sin A sin 02' 42" 6-8950899 = 0-0000001 6-8950898 log 690 2-8388491 constant 4-685 5749 7-5244240 = 0-0000008 7-524423 2 Summary (1) 00° 02' 42" 6-895089 8 00° 11' 30" 7-524 423 2 SURVEYING TRIGONOMETRY 77 AB = 66-803 ft AC = BC sin B sin^4 AC = 54-246 ft Logs 12-736 = 1-1050331 S/14' 12" = 2-929 2793 4-6855749 8-719887 3 -|(sec-10) 0-0000035 8-7198838 S/02' 42" 6-8950898 AB 1-8247940 12-736 sin 11' 30" sin 02' 42" 12-736 1-1050331 S/ll' 30" 7-524423 2 8-629 4563 S/02' 42" 6-895 0898 AC 1-7343665 2.3 Trigonometrical Ratios of the Sums and Differences of two angles (Fig. 2.20) To prove: sin (A + B) = sin A cos B + + cos A sin B (2.17) cos (A + B) = cos A cos B - - sin A sinS (2.18) Let the line OX trace out the angle A and then the angle B. Take a point P on the final line 0X 2 . Draw PS and PQ perpendicu- lar to OX and OX, respectively. Fig. 2.20 Through Q draw QK parallel to OX to meet PS at R. Draw QT per- pendicular to OX. ffPQ = KQO = A RS + PR sin(A + B) = fl = OP OP RS , PR OP OP = RS OQ PR PQ OQ 'OP PQ' OP = sin^4 cos 6 + cos A sinB 78 SURVEYING PROBLEMS AND SOLUTIONS cosC4 + fi) = OS = OT-ST m OT_ST OP OP OP OP OT 0£_ ST P£ OQ' OP PQ' OP = cos 4 cosB - sin 4 sinB If angle B is now considered -ve, sin(4-B) = sin 4 cos (-B) + cos 4 sin (-B) = sin 4 cosB - cos 4 sinB (2.19) Similarly, cos(4-B) = cos 4 cps(-B) - sin 4 sin(-B) = cos A cos B + sin A sin B (2.20) tanG4'+B) = s * n (A + ff) _ sin A cosB + cos A sinB cos (4 + B) cos A cos B - sin4 sin B (-by cos4 cosB) = tan 4 + tanB (2 . 2 1) 1 - tan A tan B Similarly, letting B be -ve, tanC4-B) = tan 4 + tan(-B) l-tan4 tan(-B) tan A - tanB 1 + tan4tanB (2.22) If sin(4 + B) = sin A cosB + cos 4 sinB and sin (A - B) = sin A cosB - cos A sinB then sin (A + B) + sin(4 - B) = 2 sin A cosB (2.23) and sin(4 + B) - sin (A - B) = 2 cos A sinB (2.24) Similarly, as cos (4 + B) = cos 4 cosB - sin 4 sinB and cos(4 - B) =• cos 4 cos B + sin 4 sin B cos(4 + B) + cos(4 - B) = 2 cos 4 cosB (2.25) cos(4 + B) - cos(4 - B) = -2 sin 4 sinB (2.26) If A « B, then sin (4 + 4) = sin 2A = 2 sin4 cosA (2.27) cos(4 + 4) = cos 24 = cos 2 4 - sin 2 4 = 1 - 2 sin 2 4 (2.28) or =2 cos 2 4 - 1 (2.29) SURVEYING TRIGONOMETRY 2 tan A tan04 + 4) = tan 2 A = 1 - tan 2 A From if and then Similarly, From 2.4 Transformation of Products and Sums sin(4 + B) + sin(A - B) = 2 sin A cosB sin 04 + B) - sinC4 - B) = 2 sinB cos 4 A + B = C A - B = D A = C-±D and B = £^_D cos(v4 + B) + cos(A - B) = 2 cos ,4 cosB cos04 + B) - cos(A - B) = -2 sinA sinfi C - D cos C + cos D = 2 cos c + D cos and cosC - cosD = -2 sin C + D sin C ~ P 2 2__ These relationships may thus be tabulated : sin (A ± B) = sin A cos B ± cos A sin B cos 04 ± B) = cos /4 cos B + sin A sin B tan 04 ±B) tan>l ± tanB 79 (2.30) 2 "" " 2 sin C + sinD = 2 sin C + D cos C ~ D (2.31) sinC sinD - 2 sin 6 " ~ D cos C + D 2 2 (2.32) (2.33) (2.34) 1 T tan 4 tan B sin04 + B) + sin 04 - B) = 2 sin A cosB sin04 + B) - sin (4 - B) = 2 cos A sinB cos(i4 + B) + cos(A - B) = 2 cos 4 cos B cos(/4 + B) - cos(/4 -B) = -2sini4sinB sin 2,4 = 2 sin ,4 cos /} cos 2A = cos 2 A - sin 2 A = 1 - 2sin 2 4 = 2 cos 2 ,4 - 1 tan 24 = 2tan ^ 1 - tanM sin A + sinB 2 sin 4 + B„^A - B cos 80 SURVEYING PROBLEMS AND SOLUTIONS sin A - sin B = 2 sin' 4 ~ B cos^ 4 + B 2 2 cos 4 + cos £ = 2 cos ^ + B cos ^ ~ g 2 2 cos ,4 - cos B = -2 sin ^ + B sin ^ ~ B 2 2 2.5 The Solution of Triangles The following important formulae are now proved: Sine rule . a = b = _£_ = 2R (2.35) sin A sinS sinC Cosine rule c z = a 2 + b z - 2ab cosC (2.36) sinC = 2l^\s(s -a){s - b){s - c)\ (2.37) ab Area of triangle = lab sinC (2.38) = ^s(s-a)(s-b)(s-c) (2.39) Half-angle formulae sini = f (s-b)(s-c) (2 .40) 2 V be tan Napier's tangent rule B -C an — — -l = b + c A = A(s-a) (2.41) 2 W be A = As-b)(s-c) (2.42) 2 V s ( s _ a) tan*^ = An£tan£±^ ( 2>43) 2.51 Sine rule (Figs. 2.21 and 2.22) Let triangle /4SC be drawn with circumscribing circle. Let 4J3, be a diameter through /I (angle ABC = angle 4^0. AC In Fig. 2.21, j£- = sinB In Fig. 2.22, i£ = sin(180-B) AB | = sinB JL = sinB 2R SURVEYING TRIGONOMETRY 81 Fig. 2.21 Fig. 2.22 Similarly sin A b sinB b sinB = 2R sinC = 2R (2.35) 2.52 Cosine rule (Fig. 2.23) AB 2 = Obtuse Fig. 2. 23 The cosine rule .2 „„ 2 Acute' with C acute AD 2 + BD 2 (Pythagoras) = AD 2 + (BC - CD) 2 = b 2 sin 2 C + {BC - b cos Cf = AD 2 + (BC + CD) 2 = 6 2 sin 2 (180 - O + {BC + b cos (180 - C)\ 2 } with C obtuse = b 2 sin 2 C + (BC - b cos C) 2 .'. AB 2 = b 2 sin z C + (BC-bcosC) 2 in either case = b 2 sin 2 C + a z - 2ab cosC + b z cos z C. = a 2 + b 2 (sin 2 C + cos 2 C) - lab cosC = a 2 + b 2 - lab cos C (2.36) 82 SURVEYING PROBLEMS AND SOLUTIONS 2.53 Area of a triangle From c 2 = a 2 + b 2 - lab cosC cos C = ~2 u z ~ z a + b - c sin 2 C = 1 - cos 2 C = 1 _( a Z +b 2 -c 2 \ 2 \ lab / lab lab = (l + a z +b 2 -c% a 2 + b 2 -c 2 ) V lab A lab J _ (a + b) 2 - c 2 x c 2 - (a - b) 2 2ab 2afe (a + b + c)(- c + a + bXc - a + b)(c + a - fc) (2ab) 2 4s (s - a) (s - b)(s - c) a 2 b z where 2s = a + b + c. ab*j sinC = i_ /s(s-a)(s-b)(s-c) (2.37) In Fig. 2.23, Area of triangle = \AD . EC = lab sin C (2.38) = ^Lab -4 V s ( s - a)(s - b)(s - c) = ^s(s-a)(s-b)(s-c) (2.39) 2.54 Half-angle formulae From Eq.(2.28), tint* A -%1-co.A) . ^(l-»Lt£L-L) q 2 _ (fc _ C ) 2 4bC (a - b + c)(a + b - c) Abe = (s-b)(s-c) be sini = Ks-bKs-e) (2 40) 2 V fee SURVEYING TRIGONOMETRY 83 Similarly, cosz±A = I(l + cos4) = l(l + b 2 +c 2 -a 2 \ 2 2 2\ 2bc * = (b + c) 2 - a 2 4bc = (b + c + a)(b + c - a) Abe _ s(s - a) be cosil = /sjs-a) (241) 2 V be tan_d = . A sin cos 2 - = / (s-bXs-c) (2>4 2) A V s(s - a) The last formula is preferred as (s - b) + (s - c) + (s - a) = s, which provides an arithmetical check. 2.55 Napier' s tangent rule From the sine rule, then b _ sinB c sinC b - c sinS - sinC b + c sinS + sinC 2cos S + C sin B - C 2 2 r sin B + C ™c B-C 2 2 tan B ~ C 2 B + C tan (By Eqs. 2.31/2.32) •*. tan B ~ c = b-c tQV B + C (2.43) 2 b + c 2 2.56 Problems involving the solution of triangles All problems come within the following four cases: 84 SURVEYING PROBLEMS AND SOLUTIONS (1) Given two sides and one angle (not included) to find the other angles. Solution : Sine rule solution ambiguous as illustrated in Fig. 2.24. Given AB and AC with angle B, AC may cut line BC at C, or C 2 Fig. 2.24 The ambiguous case of the sine rule (2) Given all the angles and one side to find all the other sides. Solution: Sine rule (3) Given two sides and the included angle Solution: Either cos rule to find remaining side or Napier's tangent rule (this is generally pre- ferred using logs) (4) Given the three sides Solution: either cos rule or half-angle formula (this is generally preferred using logs.) Example 2.8 (Problem 1) Let c = 466-0 m a = 190-5 m A = 22° 15' Using sine rule, sin C sin A c a sinC = csinA a = 466-0 sin 22° 15' 190-5 466-0 x 0-378 65 190-5 = 0-926 25 Fig. 2.25 SURVEYING TRIGONOMETRY 85 Angle C 2 = 67° 51' 30" or 180 C, = 112 08' 30" To find side b (this is now Problem 2) , a sin B sin A fo _ a sinB sin ,4 i.e. 5, = Log calculation 190-5 sin [180 -(67° 51' 30" +22° 15' 00")] sin 22° 15' 00" 190-5 sin89°53'30" sin 22° 15' 00" = 190 ' 5 x 1-0 = 503-10 m 0-37865 190-5 sin[(180 -(112° 08' 30"+ 22° 15' 00")] sin 22° 15 '00" 190-5 sin 45° 36' 3 0" = 359 . 51m sin 22° 15' 00 log sin C = logc + log sin A - log a C = 67° 51' 30" 466-0 ©2-668 39 sin 22° 15' 9-578 24 ©2-24663 190-5 2-279 90 sinC ©9-96673 N.B. The notation 9-578 24 is preferred to 1-578 24 - this is the form used in Chambers, Vegas, and Shortredes Tables. Every characteristic is increased by 10 so that subtraction is sim- plified — the ringed figures are not usually entered. Also logfc = log a + log sin B + log cosec,4 "I.B. Addition using log cosec4 is preferable to subtracting log sin A. e. logfc, = logl90-5 + log sin 89° 53' 30" + log cosec22° 15'00" 190-5 2-27990 sin89°53'30' 0-0 cosec22°15'00" 10-42176 fr, = 503-10 m b, 2-70166 86 SURVEYING PROBLEMS AND SOLUTIONS log& 2 = logl90-5 + log sin 45° 36' 30 " + log cosec22° 15'00" 190-5 2-27990 sin45°36'30" 9-85405 cosec 22° 15' 00" 10-421 76 b z = 359-51 m b 2 2-55571 N.B. A gap is left between the third and fourth figures of the logarithms to help in the addition process, or it is still better to use squared paper. Example 2.9 (Problem 3) Let a = 636 m c = 818 m B = 97° 30' To find b, A and C. By cosine rule b 2 = a* + c 2 - 2ac cosB = 636 2 + 818 2 - 2 x 636 x 818 x cos = 404496 + 669124 + 135815-94 = 1209435-94 b . 1099-74 m sin A sin B a b oinA asinB 636 sin 97° 30' SmA = b = 1099-74 = 0-57337 A = 34° 59' 10" B = 47° 30' 50" The first part of the calculation is essentially simple but as the figures get large it becomes more difficult to apply and logs are not suitable. The following approach is therefore recommended. As c> a, C> A. Then, by Eq.(2.43), tan%4 , £^JltanC±J. 2 c + a 2 818 - 636 . (180 -97° 30') 818 + 636 2 182 tan 41° 15' 1454 = 0-10977 c = 47° 30' 50" A = 34° 59' 10" b = a sin 6 cosec A SURVEYING TRIGONOMETRY 87 \{C-A) = 6° 15' 50" l -(C + A) = 41° 15' 00" By adding By subtracting Now, by sine rule, = 636 sin 97° 30' cosec 34° 59' 10" = 1099-74 m N.B. This solution is fully logarithmic and thus generally preferred. Also it does not require the extraction of a square root and is there- fore superior for machine calculation. Example 2.10 (Problem 4) Let a = 381 b = 719 c = 932 To find the angles. a 2 cos A From then b 2 + c 2 - 2bc cos A b 2 + c 2 - a 2 2bc 719 2 + 932 2 - 381 2 2 x 719 x 932 516 961 + 868 624 - 145 161 1 340 216 A = 0-925 54 22° 15' 00" cosfi 2 „2 t 2 a + c - b 2ac 145161 + 868 624 - 516961 2 x 381 x 932 496824 710 184 B = 45° 36' 30 = 0-69957 cosC = a 2 + b 2 - c 2 lab 145161 + 516961 - 868624 2 x 381 x 719 -206 502 547 878 = -0-37691 88 SURVEYING PROBLEMS AND SOLUTIONS C = 180 - 67° 51' 30" = 112° 08' 30" Check 22° 15' 00" + 45° 36' 30" + 112° 08' 30" = 180° 00' 00" Alternative By half-angle formula, tan S. = Ks - b)(s - c) 2 V s(s - a) a = 381 s - a = 635 b = 719 s - b = 297 c = 932 s - c = 84 2s =2032 s = 1016 s = 1016 then tanA - / 297 * 84 2 V 1016 x 635 This is best solved by logs log tani = I[(log297 + log84) - (log 1016 + log635)] 2 2 = 11° 7' 30" = 22° 15' 00" 297 2-47276 84 1-924 28 tan^ = / 635 >< 84 2 V1016 x 297 -^ = 22° 48' 15" 2 B = 45° 36' 30" 4-39704 1016 3-00689 635 2) 1/2 2-80277 5-80966 18-58738 tany 9-29369 635 2-80277 84 1-924 28 4-727 05 1016 3-00689 297 2-47276 5-47965 2) 19-24740 tan B/2 9-623 70 Example 2.11 SURVEYING TRIGONOMETRY 89 The sides of a triangle ABC measure as follows: AB = 36ft Olin., AC = 30 ft if in. and BC = 6ft0lin. lo 4 4 (a) Calculate to the nearest 20 seconds, the angle BAC. (b) Assuming that the probable error in measuring any of the sides is ± 1/32 in. give an estimate of the probable error in the angle A. (M.Q.B/S) Fig. 2. 26 AB = c = 36 ft 0* in. = 36-036 ft 16 AC = b = 30ftlf-in. = 30-146 ft 4 S - C = BC = a = Using Eq. (2.42) 6ft0±in. = 4 6-021 ft 2 |72-203 s = 36-1015 c = 0-0655 5-9555 s - a = 30-0805 s = 36-1015 tan-d = /Cs-bHs-c) 2 V s(s-a) J, 5-955 5x0-065 5 36*1015x30-0805 4- = 1°05'10" 2 A = 2 o 10 , 20 /, tan# = / (s-o)(s-"c) 2 V s(s - 6) y 30-0805x0-065 5 36-1015x5-9555 ^ = 5° 28' 05" 2 g = 10° 56 '10" tan£ = Rs-a)(s-b) 2 V s(s-c) ■J 30-0805x5-9555 36-1015x0-0655 90 SURVEYING PROBLEMS AND SOLUTIONS £. = 83° 26' 45" 2 C = 166° 53' 30" Check A + B + C = 180° (b) The probable error of ±1/32 in. = ±0*003 ft. The effect on the angle A of varying the three sides is best cal- culated by varying each of the sides in turn whilst the remaining two sides are held constant. To carry out this process, the equation must be successively differentiated and a better equation for this purpose is the cosine rule. b z + c 2 - a 2 Thus cos A = Differentiating with respect to a, -sin A8A a = 8A„ = 2bc 2a 8a 2bc a 8a be sin A Differentiating with respect to b, -sin A8A h = (2bcx2b)-(b 2 +C 2-a*)(2c) 4b*c'- but cos C X c 2b 2 c a 2 + b - c 2 2b 2 c a + b - c 2ab _ :osC * 6 be sin A 8A b = - a cos C 8b = - 8A a cos C (as 8a = 8b) Similarly, from the symmetry of the function: 8A C = _ « cos g g c = SA a cos B (as 8a = 8c) be sin A Substituting values into the equations gives: 8A = 6-021 x ±0-003 x 206265 ±90-49 sec 30-146 x 36-036 x sin 2° 10' 20" 8A b = S4 cos 166° 53' 30" = ±88-13 sec 8A„ = &4„cos 10° 56' 10" = ±88-85 sec SURVEYING TRIGONOMETRY 91 Total probable error = V&4 2 . + 5-4 2 + 8A 2 C = V90-49 2 + 88-13 2 + 88-85 : = ± 154 seconds. 2.6 Heights and Distances 2.61 To find the height of an object having a vertical face The ground may be (a) level or (b) sloping up or down from the observer, (a) Level ground (Fig. 2.27) Fig. 2.27 V.D. The observer of height h is a horizontal distance (H.D.) away from the object. The vertical angle (V.A.) = 6 is measured. The vertical difference V.D. = H.D. tanfl (2.44) Height of the object above the ground = V.D. + h V.D. V.D. \B Horizontal line ._o< Depression" H.D Fig. 2-28 (b) Sloping ground (Fig. 2.28) The ground slope is measured as a V.D. = H.D. (tan0 ± tana) (2.45) Height of object above the ground = V.D. + ft N.B. This assumes that the horizontal distance can be measured. 92 SURVEYING PROBLEMS AND SOLUTIONS 2.62 To find the height of an object when its base is inaccessible A base line must be measured and angles are measured from its extremities. (a) Base line AB level and in line with object. (Fig. 2.29) Vertical angles a and /8 are measured. Fig. 2 29 A,C, = EC, cot a B,C, = EC, cot/8 A,B, = A,C, - B,C, = EC, (cot a -cot 0) Thus EC, = AB cot a - cot j8 Height of object above ground at A = EC, + ft, Height of ground at D above ground at A = EC, + h, — h 2 (2.46) (b) Base line AB level but not in line with object (Fig. 2.30) Angles measured at A horizontal angle vertical angle a at B horizontal angle <fi vertical angle /8 In triangle ABC ; AC = AB sin<£ cosec(0 + 0) (sine rule) SURVEYING TRIGONOMETRY 93 Fig. 2.30 and BC = AB sin 6 cosec(0 + <£) Then C,£ = 4Ctana = ,46 sin cosec(0 + <£) tana (2.47) Also C,£ = BC tan/8 = 4B sin0 cosec(0 + 0) tan£ (2.48) Height of object (E) above ground at A = C,£ + A, Height of ground (D) above ground at A = C,E + /i, - /i 2 (c) Sase Zme 4B on sloping ground and in line with object (Fig. 2.31) Fig. 2.31 94 SURVEYING PROBLEMS AND SOLUTIONS Angles measured at A: vertical angle a (to object) vertical angle 8 (slope of ground) Angle measured at B : vertical angle j8 (to object) In triangle A^EB X , 4, = a - 8 B, = 180 - (£ - 8) E = p - a Then A,E = A,B, sin {180 -(0-S)} cosec(/8 - a) = AB sin(/8 - 8) cosecQ8 - a) (2.49) Height of object (£) above ground at A EC = EC, + h } = A^E sin a + /i, = AB sin(fi - 8) cosec(/8 - a) sina + /i, (2.50) Height of ground (D) above ground at A = EC, + hy — h 2 . (d) itase line AB on sloping ground and not in line with object (Fig.2.32) Fig. 2.32 Angles measured at A : horizontal angle 6 vertical angles a (to object) 8 (slope of ground) at B: horizontal angle <f> vertical angle /S A y B 2 = AB cos 8 SURVEYING TRIGONOMETRY 95 Then 4,0, = A X B 2 sin<£cosec(0 + <f>) EC, = 4,C, tana — AB cos 8 sin0 cosec(fl + <ft) tana (2.51) Height of object (E) above ground at A EC = EC X + hi E,C 2 = B 2 C, = y4,B 2 sin0 cosec(0 + <£) EC 2 = B,C 2 tan/3 Similarly, height of object (E) above ground at j4 EC = EC 2 + fi,B 2 + h, = 4g cos 5 sing cosec(fl + <f>) tan/6 + AB sing + ft, ,~ co\ Height of ground (D) above ground at 4 = EC\ + hi — h z or = EC 2 + h, - h 2 + AB sinS 2.63 To find the height of an object above the ground when its base and top are visible but not accessible (a) Base line AB, horizontal and in line with object (Fig. 2.23) Vertical angles measured at A : a, and a 2 at B: /S, and B z Fig. 2.33 96 SURVEYING PROBLEMS AND SOLUTIONS From Eq.(2.46) Also C,E = C,D = AB cot a, - cot/3, AB cot a 2 - cotB z Then ED = C,E - C,D = H H = AB\ [cota, - cot^3, cota 2 - cotB z J (b) Base line AB horizontal but not in line with object (Fig. 2.34) Angles measured at A : horizontal angle 6 vertical angles a, and a 2 at B : horizontal angle <f> vertical angles jS, and B z (2.53) I MHEWM^WAW/SigWig Fig. 2. 34 AC = 4,C, ED = H .'. H Similarly, H = AB sin0 cosec(# + <f>) = AC (tan a, - tan a 2 ) = AB sin <f> cosec(0 + ^>)(tana, - tana 2 ) (2.54) = AB sin 6 cosec (0 + <£) (tan B , - tan /8 2 ) (2 . 55) SURVEYING TRIGONOMETRY 97 (c) Base line AB on sloping ground and in line with object (Fig. 2.35) Vertical angles measured at A : a,, a 2 and 8 at B: /8, and /8 2 Fig. 2.35 From Eq.(2.50) EC, = /IB sin(/3, - 8) cosec(/9, - a,) sin a, and DC, = 4B sin(/3 2 - 5) cosecQ3 2 - a 2 ) sin a 2 Then ED = EC, - DC, H = A8[sin(/6, - 5) cosec(/3, - a,) sin a, - - sin(/8 2 - 8) cosec(j8 2 - a 2 ) sin a 2 ] (2.56) (d) Base line AB on sloping ground and not in line with object (Fig. 2.36) Angles measured at A: horizontal angle 6 vertical angles a, and a 2 8 (slope of ground) at B: horizontal angle cf> vertical angles /3, and £ 2 98 SURVEYING PROBLEMS AND SOLUTIONS Fig. 2. 36 From Eq.(2.51), ECi = AB cosS sin<£ cosec(0 + <f>) tana, DC X = AB cos 8 sin<£ cosec(<9 + <f>) tana 2 Then ED = EC, - DC, = AB cos 5 sin0 cosec(0 + <£)(tana, - tana 2 ) (2.57) Also ED = AB cos 5 sin 6 cosec(<9 + <£)(tan)6, - tan B z ) (2.58) 2.64 To find the length of an inclined object (e.g. an inclined flagstaff) on the top of a building (Fig. 2.37) Base line AB is measured and if on sloping ground reduced to horizontal. Angles measured at A: horizontal angles 0, and d z to top and bottom of pole vertical angles a, and a 2 to top and bottom of pole at B: horizontal angles <£, and <f> 2 to top and bottom of pole vertical angles )6, and B z to top and bottom of pole. In plan the length ED is projected as E^D^ (= E Z D). In elevation the length ED is projected as EE 2 , i.e. the difference in height Then AE, = AB sin</>, cosec(0, + <£,) Also AD, = AB sin 2 cosec(0 2 + </> 2 ) SURVEYING TRIGONOMETRY 99 e d Fig. 2.37 The length E,D = L is now best calculated using co-ordinates (see Chapter 3) Assuming bearing AB = 180° 00' Ae = AE, sin(90 -6,) = AE, cos0, 4d = 4D, sin(90 - 2 ) = 4D, cos0 2 Then ed = D 2 D, = /Id - 4e = AD, cos6 2 - 4E, cos0, and similarly, E,D 2 = E,e _ D 2 e = 4£, sin0, - AD, sin0 2 In the triangle E,D,D 2 . The bearing of the direction of _ tan -i D 2 D, inclination (relative to AB) E,D 2 Length E,D, = E,D 2 (sec bearing) To find difference in height E E 2 Height of top above A = AE, tana, Height of base above A = AD, tan a 2 Length EE 2 = AE, tana, - AD, tan a 2 To find length of pole : In triangle EDE 2 , i.e. ED 2 ED EE\ + E 2 D Z -V EEl + E,D 2 100 SURVEYING PROBLEMS AND SOLUTIONS 2.65 To find the height of an object from three angles of elevation only (Fig. 2.38) Fig. 2.38 Solving triangles ADB and ADC by the cosine rule, cos* = h 2 cot 2 a + x z -h 2 cot z p 2hx cot a fc 2 cot 2 a + (x + y) 2 - h 2 cot 2 (2.59) 2h(x + y) cot a ••• (x+y)[fc 2 (cot 2 a - cot 2 j8) + x 2 ] = x[h\cot 2 a - cot 2 0) + (x + y) 2 ] i.e. h 2 [(x + y)(cot 2 a - cot 2 j9) - x(cot 2 a - cot 2 0)] = x(x + y) 2 - x 2 (x + y) i.e. h 2 = (x + y)[x(x + y) - x 2 ] (x + y)(cot 2 a - cot 2 /S) - x(cot 2 a - cot 2 0) (x + y)(xy) (x + y)(cot 2 a - cot 2 j8) - x(cot 2 a - cot 2 0) [ xy(x + y) ] [x(cot 2 - cot 2 /8) + y(cot 2 a - cot 2 j8)J H x = y, h = y/2x [cot 2 6 - 2 cot 2 /3 + cot 2 a]* (2.60) (2.61) SURVEYING TRIGONOMETRY 101 Example 2.12 A,B and C are stations on a straight level line of bearing 126° 03' 34". The distance AB is 523-54 ft and BC is 420-97 ft. With an instrument of constant height 4'- 3" vertical angles were successively measured to an inaccessible up-station D as follows: At A 7° 14' 00" B 10° 15' 20" C 13° 12' 30" Calculate (a) the height of station D above the line ABC (b) the bearing of the line AD (c) the horizontal length AD. (R.I.C.S.) D Fig. 2.39 (a) In Fig. 2.39, AD = h cot a BD = h cot/8 CD = h cot0 Solving triangles AD % B and AD X C, using Eq. (2.60), _ i 1 2 h = xy(x + y) *(cot 2 - cot 2 /9) + y(cot 2 a - cot 2 /S) (523-54 x 420-97)(523-54 + 420-97) 523-54(cot 2 13° 12' 30 - cot 2 10° 15' 20") + 420-97(cot 2 7° 14' 00" - cot 2 10 = 175-16 ft — 1 '15'20"J 102 SURVEYING PROBLEMS AND SOLUTIONS .*• Difference in height of D above ground at A = 175-16 + 4-25 = 179-41 ft Using Eq. (2. 59), ™,a h\cot 2 a - cot 2 fi) + x 2 COS <p = * 7T- — ^ ^ 2hx cot a 175-16 2 (cot 2 7°14'00 - cot 2 10° 15' 20") + 523-54 s 2 x 175-16 x 523-54 x cot 7° 14' 00" = 0-85909 <f> = 30° 47' 10" (b) Thus bearing of AD = 126° 03' 34' - 30° 47' 10" = 095° 16' 24" (c) Length of line AD y = h cot a = 175-16 cot 7° 14' = 1380-07 ft 2.66 The broken base line problem Where a base line AD cannot be measured due to some obstacle the following system may be adopted, Fig. 2.40. Fig. 2.40 Broken base line Lengths x and z are measured. Angles a, j8 and 6 are measured at station E. To calculate BC = y : Method 1 In triangle AEB EB = In triangle AEC EC = x sin EAB sin a (x + y) sin EAC sin (a + /S) SURVEYING TRIGONOMETRY 103 Then S? = * sin ( a + ft) (2.62) EC (x + y) sin a Also in triangle EDB EB = Cv + z) sin EDB sin(0 + ft) in triangle EDC EC = z sinEDC sin# Then ER = fr + z > sin( ^ (2.63) EC zsin(0+ft) Equating Eqs (2.62) and (2.63) (y + z) sinfl _ x sin (a + ft) z sin (0 + ft) (x + y) sin a i.e. (x + yXy + z) = * z sin (a + ft) sin(fl + ft) sin a sin# Then yz + y( x + z) + xz 1 - sin ( a + ft) sin ((9 + ft) sin a sin# = (2.64) This is a quadratic equation in y. Thus y = _ x + z + //*_+_z\ 2 _ ^[l _ sin(q + ft) sin(fl + ft) 1 2 /J \ 2 J I sin a sin0 J ±£+ /(* - Z ) 2 + r 7 Si "( a + ^ Sin (0 7ft) (265) 2 V V 2 / sin a sin# Method 2 Area of triangle 4 BE = Axfc = I^E.EB sin a (1) BCE = iyfc = IfiE. EC sin ft (2) CUE = ±zh = ICE. ED sin (9 (3) ADE = |(jc + y + z)h = ±AE . ED sin(a + ft + 0) (4) Dividing (1) by (2) x AE sin a y EC sin ft (5) Dividing (3) by (4), Z ~ - AB CE f Sin * _ (6) 6 JV x + y + z AEsin(a + 8 + 6) 104 SURVEYING PROBLEMS AND SOLUTIONS Multiplying (5) by (6), xz sin a sin0 y(x + y + z) sin/3 sin(a + j8 + d) i.e. y* + y(x + z) - xz sin/3sin(a + /3 + fl) = sin a sin# Then y = -(*±±) + /fc±£V + xz sin P sin < a + Z 3 + ^> (2 \ 2 / sj\ 2 / sina sinfl v ' Method 3 (Macaw's Method) In order to provide a logarithmic solution an auxiliary angle is used. From the quadratic equation previously formed, 66) y 2 + y(x + z)- xz sin/3sin(a + /3 + fl) = (2 67) sina sin# i.e. {y + jfc, + z)P - J(, + zY + " Sin f in s ^°, + / + g) (2-68) Now let tan'M = 4«z si„ fl gin (a + p + (?) (2 69) (x + z) sin a sin Substituting this in Eq. (2.68), we get { y + |(x + z)} 2 = I(x + z) 2 (l + tan 2 M) = i(jc + z) 2 sec 2 iW y + I(x + z) = |(x + z) secM y = I(x + z)(secM - 1) = (x + z) sec/Wkl - cosM) y = (x + z) secM sin 2 iiW (2.70) Example 2.13 The measurement of a base line AD is interrupted by an obstacle. To overcome this difficulty two points B and C were established on the line AD and observations made to them from a station E as follows: AEB = 20° 18' 20" BEC = 45° 19' 40" CED = 33° 24' 20" Length AB = 527-43 ft and CD = 685-29 ft. Calculate the length of the line AD. (R.I. C.S.) SURVEYING TRIGONOMETRY 105 Here a = 20° 18' 20" \ a + p = 65° 38' 00" p = 45°19'40" a + p + d = 99°02'20" = 33° 24' 20" j p + d = 78° 44' 00" x = 527-43 | \{x + z) = 1(1212-72) = 606-36 z = 685-29 J |(jc^z) = §(157-86) = 78-93 By method 1 y = -606-36 + As .93' , 527-43 x 685-29 sin 65° 38' sin 78° 44' V sin20°18'20"sin33°24'20" = -606-36 + V6230 + 1690057 = -606-36 + 1302-415 = 696-055 ft Then AD = 1212-72 + 696-055 = 1908 -775 ft By method 2 , y = -606-36 + yfa)6-36 2 + 527 ' 43 x 685 ' 29 sin 45°19'40" sin99°02'20" ^ sin 20° 18' 20" sin 33° 24' 20" = -606-36 + V367672 + 1328 614 = -606-36 + 1302-415 = 696-055 ft By method 3 By logs 4 X z sinp 0-6020600 2-7221648 2-8358744 9-851955 4 sin (a + + 0) 9-994 573 1 cosec a 10-459 637 2 cosec 10-2591940 6-725 458 9 (x + z) 2 6-1675210 tanzM 0-557 9379 taniW 0-278 968 9 _ M = 62° 15' 11" §-M = 31°07'36" 106 SURVEYING PROBLEMS AND SOLUTIONS secM 0-332017 5 sin~M 9-713433 2 sin ^M 2 9-713 433 2 (*+ z) 3-083760 5 2-842644 4 y = 696-055 2.67 To find the relationship between angles in the horizontal and inclined planes (Fig. 2.41) Fig. 2.41 Let (1) lines /4B, and B,C be inclined to the horizontal plane by a and /S respectively. (2) Horizontal angle ABC = 6 (3) Angle in inclined plane AB^C = <f> (4) B,B = h Then AB = h cot a AB, = h coseca BC = h cot/8 B,C = fccosec/3 In triangle ABC, 4C 2 = AB 2 + BC 2 - 2 AB.BC cos (9 = h 2 cot 2 a+ h 2 cot 2 fi - 2 A 2 cot a cot /8 cos Similarly in triangle AB,C, AC 2 = h 2 cosec 2 a + /z 2 cosec 2 /3 - 2fc 2 coseca cosec/S cos<£ SURVEYING TRIGONOMETRY 107 Then h z cot 2 a + /i 2 cot 2 ft -2fc 2 cotacotft cos0 = /i 2 cosec 2 a + h 2 cosec 2 B - 2h z coseca cosecft cos<£ cos *0 = (cosec 2 a - cot 2 a) + (cosec 2 ft - cot 2 ft) + 2 cot a cot ft cosfl 2 coseca cosecft cosec 2 a - cot 2 a = cosec 2 ft - cot 2 ft = 1 as Then cos<£ = 2(1 + cot a cot ft cosfl) 2 coseca cosecft or cos = sin a sin/8 + cos a cos ft cos# (2.71) 6 = cos - sin a sin ft ,n y<y\ cos a cos ft Example 2.14 From a station 4 observations were made to stations B and C with a sextant and an abney level. With sextant - angle BAC = 84° 30' With abney level - angle of depression (AB) 8° 20' angle of elevation (AC) 10° 40' Calculate the horizontal angle BA^ C which would have been measured if a theodolite had been used (R.I.C.S./M) From equation (2.72), cos0 = cos - sin a sin ft cos a cos ft = cos 84° 30' - sin (-8° 20') sin 10° 40' cos (-8° 20') cos 10° 40' cos 84° 30' + sin8°20'sinl0°40' cos 8° 20' cos 10° 40' 0-09585 + 0-14493 x 0-18509 0-989 44 x 0-98272 0-12268 0-97234 = 0-12617 d = 82° 45' 10" Example 2. 15 A pipe-line is to be laid along a bend in a mine roadway ABC. If AB falls at a gradient of 1 in 2 in a direction 036° 27', whilst BC rises due South at 1 in 3-5, calculate the angle of bend in the pipe. (R.I.C.S.) 108 SURVEYING PROBLEMS AND SOLUTIONS 3.5 Fig. 2.42 Plan From equation (2.71), cos0 = sin a sinjS + cos a cos/3 cos# where a = cot" 1 2 = 26° 33' £ = cot" 1 3-5 = 15° 57' 6 = 036° 27' - 00° = 36° 27' .-. cos0 = sin 26° 33' sin 15° 57'+ cos 26° 33' cos 15° 57' cos 36° 27' cf> = 35° 26' 40" i.e. 35° 27' Exercises 2(b) 8. Show that for small angles of slope the difference between hori- zontal and sloping lengths is h z /2l (where h is the difference of verti- cal height of the two ends of a line of sloping length I) If errors in chaining are not to exceed 1 part in 1000, what is the greatest slope that can be ignored ? [L.U/E Ans. 2° 34'] 9. The height of an electricity pylon relative to two stations A and B (at the same level) is to be calculated from the data given below. Find the height from the two stations if at both stations the height of the theodolite axis is 5'— 0". SURVEYING TRIGONOMETRY 109 Pylon Data: AB = 200 ft Horizontal angle at A = 75° 10' at B = 40° 20' Vertical angle at A = 43° 12' at B = 32° 13' A " ' x fl (R.I.C.S. Ans. mean height 139-8 ft) 10. X, Y and Z are three points on a straight survey line such that XY = 56 ft and YZ = 80 ft. From X, a normal offset was measured to a point A and X/4 was found to be 42ft. From Y and Z respectively, a pair of oblique offsets were measured to a point B, and these distances were as follows: YB = 96 ft, ZB = 88 ft Calculate the distance 4fi, and check your answer by plotting to some suitable scale, and state the scale used. (E.M.E.U. Ans. 112-7 ft) 11. From the top of a tower 120 ft high, the angle of depression of a point A is 15°, and of another point B is 11°. The bearings of A and B from the tower are 205° and 137° respectively. If A and B lie in a horizontal plane through the base of the tower, calculate the distance AB. (R.I.C.S. Ans. 612 ft) 12 A, 6, C, D are four successive milestones on a straight horizon- tal road. From a point due W of A, the direction of B is 84°, and of D is 77°. The milestone C cannot be seen from 0, owing to trees. If the direction in which the road runs from A to D is 0, calculate 0, and the distance of from the road. (R.I.C.S. Ans. = 60°06' 50", OA = 3-8738 miles) 13. At a point A, a man observes the elevation of the top of a tower B to be 42° 15'. He walks 200 yards up a uniform slope of elevation 12° directly towards the tower, and then finds that the elevation of B has increased by 23° 09'. Calculate the height of B above the level of A. (R.I.C.S. Ans. 823-82 ft) 14. At two points, 500 yards apart on a horizontal plane, observations of the bearing and elevation of an aeroplane are taken simultaneously. At one point the bearing is 041° and the elevation is 24°, and at an- other point the bearing is 032° and the elevation is 16°. Calculate the height of the aeroplane above the plane. (R.I.C.S. Ans. 1139ft) 110 SURVEYING PROBLEMS AND SOLUTIONS 15. Three survey stations X, Y and Z lie in one straight line on the same plane. A series of angles of elevation is taken to the top of a colliery chimney, which lies to one side of the line XYZ. The angles measured at X, Y and Z were: at X, 14° 02'; at Y, 26° 34'; at Z, 18° 26' The lengths XY and YZ are 400 ft and 240 ft respectively. Calculate the height of the chimney above station X. (E.M.E.U. Ans. 112-0 ft) 16. The altitude of a mountain, observed at the end A of a base line AB of 2992-5 m, was 19° 42' and the horizontal angles at A and B were 127° 54' and 33° 09' respectively. Find the height of the mountain. (Ans. 1804 m) 17. It is required to determine the distance between two inaccessible points A and B by observations from two stations C and D, 1000 m apart. The angular measurements give ACB = 47°, BCD = 58°, BDA = 49°; ADC = 59°. Calculate the distance AB (Ans. 2907 -4 m) 18. An aeroplane is observed simultaneously from two points A and B at the same level, A being a distance (c) due north of B. From A the aeroplane is S^°E and from B N <f> E. Show that the height of the aeroplane is c tan a sin <f> sin(0 + <f>) and find its elevation from B. /L.U. Ans. B = tan-' sin<fttana \ V sintf / 19. A straight base line ABCD is sited such that a portion of BC cannot be measured directly. If AB is 575-64 ft and CD is 728-56 ft and the angles measured from station to one side of ABCD are DOC = 56° 40' 30" COB = 40° 32' 00" BOA = 35° 56' 30" Calculate the length BC. (E.M.E.U. Ans. 259-32 ft) 20. It is proposed to lay a line of pipes of large diameter along a roadway of which the gradient changes from a rise of 30° to a fall of 10° coincident with a bend in the roadway from a bearing of N 22° W to N 25° E. Calculate the angle of bend in the pipe. 11M , n( on » N (Ans. ny J? ou ) SURVEYING TRIGONOMETRY 111 21. At a point A at the bottom of a hill, the elevation of the top of a tower on the hill is 51° 18'. At a point B on the side of the hill, and in the same vertical plane as A and the tower, the elevation is 71°40'. AB makes an angle 20° with the horizontal and the distance AB = 52 feet. Determine the height of the top of the tower above A. (L.U. Ans. 91-5 ft) 22. Two points, A, B on a straight horizontal road are at a distance 400 feet apart. A vertical flag-pole, 100 feet high, is at equal distances from A and B. The angle subtended by AB at the foot C of the pole (which is in the same horizontal plane as the road) is 80°. Find (i) the distance from the road to the foot of the pole, (ii) the angle subtended by AB at the top of the pole. (L.U. Ans. (i) 258-5 ft, (ii) 75° 28') Bibliography USILL, G.w. and HEARN, G., Practical Surveying (Technical Press). HADDOCK, M.H., The Basis of Mine Surveying (Chapman and Hall). LONEY, s.c, Plane Trigonometry (Cambridge University Press). blakey, J., Intermediate Pure Mathematics (Macmillan). 3 CO-ORDINATES A point in a plane may be defined by two systems: (1) Polar co-ordinates. (2) Rectangular or Cartesian co-ordinates. 3.1 Polar Co-ordinates This system involves angular and linear values, i.e. bearing and length, the former being plotted by protractor as an angle from the meridian. B (s,8) Fig. 3. 1 Polar co-ordinates A normal 6 inch protractor allows plotting to the nearest 1/4°; a cardboard protractor with parallel rule to 1/8°; whilst the special Book- ing protractor enables 01' to be plotted. The displacement of the point being plotted depends on the physi- cal length of the line on the plan, which in turn depends on the horiz- ontal projection of the ground length and the scale of the plotting. A s Fig. 3.2 Displacement due to angular error If a is the angular error, then the displacement 112 CO-ORDINATES 113 BB X = s tana and as a is small BB, ~ sa If s = 300 ft and a = 01' 00", DD 300 x 60 x 12 . , fiS ' = 206 265 inches ~ 1 inch i.e. 1 minute of arc subtends 1 inch in 100 yards, 1 second of arc subtends 1 inch in 6000 yards, i.e. 3 x /a miles. Similarly, on the metric system, if s = 100 metres and a = 01 ' 00", ♦i. dd 100x60 . then BB, = metres 206 265 = 0-0291 m, i.e. 29 mm Thus, 1 minute of arc subtends approximately 30 mm in 100m, 1 second of arc subtends approximately 1 mm in 200 m or 1cm in 2 km. A point plotted on a plan may be assumed to be 0*01 in., (0*25mm), i.e. 0'Olin. in 1 yard (0*25 mm in 1 metre) — > 1 minute of arc, 0*1 in. in 1 yard (25 mm in 1 metre) — * 10 minutes of arc. As this represents a possible plotting error on every line, it can be seen how the error may accumulate, particularly as each point is depend- ent on the preceding point. 3.11 Plotting to scale The length of the plotted line is some definite fraction of the ground length, the 'scale' chosen depending on the purpose of the plan and the size of the area. Scales may be expressed in various ways: (1) As inches (in plan) per mile, e.g. 6 in. to 1 mile. (2) As feet, or chains, per inch, e.g. 10ft to 1 inch. (3) As a representative fraction 1 in n, i.e. 1/n, e.g. 1/2500. 3.12 Conversion of the scales I insult ranracanfc ■.-— .- 40 40 inches to 1 mile — 1 inch represents — -rp: — feet lin. = 132 ft = 44 yd = 2 chn 114 SURVEYING PROBLEMS AND SOLUTIONS 1 inch to 132 ft - 1 inch represents 132 x 12 inches lin. = 1584 in. Thus the representative fraction is 1/1584. 3.13 Scales in common use Ordinance Survey Maps and Plans: Large scale: 1/500, 1/1250, 1/2500. Medium scale: 6 in. to 1 mile (1/10 560), 2y 2 in. to 1 mile (1/25000). Small scale: 2,l, 1 / 2 , 1 / 4 in. to 1 mile; 1/625000, 1/1250000. Engineering and Construction Surveys: 1/500, 1/2500, 10-50 ft to 1 inch, 1/4, V8, Vl6in. to 1ft. (See Appendix, p. 169) 3.14 Plotting accuracy Considering 0*01 in. (0*25 mm) as the size of a plotted point, the following table shows the representative value at the typical scales. O.S. Scales suggested measurement precision limit 1/500 0-01 x 500 = 5 in. 3 in. (76 mm) 1/1250 0-01 x 1250 = 12-5 in. 1ft (0 -3 m) 1/2500 0-01 x 2500 = 25-0 in. 2 ft (0-6 m) 1/10 560 0*01 x 10 560 = 105-6 in. 5 ft (1-5 m) 1/25000 0*01 x 25000 = 250-0 in. 10 ft (3-0 m) Engineering Scales 1 in. to 10 ft 0-01 x 120 = l-2in. lin. lin. to 50ft 0-01 x 600 = 6-0 in. 6 in. lin. to lchn 0-01 x 792 = 7-92 in. 6 in. or l / 2 link lin. to 2chn 0-01 x 1584 = 15-84 in. 1 ft or 1 link. 3.15 Incorrect scale problems If a scale of 1/2500 is used on a plan plotted to scale 1/1584 what conversion factor is required to (a) the scaled lengths, (b) the area computed from the scaled length? (a) On the plan 1 in. = 1584 in. whereas the scaled value shows 1 in. = 2500 in. All scaled values must be converted by a factor 1584/2500 = 0-6336. CO-ORDINATES 115 (b) All the computed areas must be multiplied by (0*6336jf 0-4014 3.2 Bearings Four meridians may be used, Fig. 3.3: 1. True or geographical north. 2. Magnetic north. 3. Grid north. 4. Arbitrary north. GN. T . N . 4 M.N M.N. T.N. 6.N. Convergence of the meridian Magnetic declination Fig. 3. 3 Meridians 3.21 True north The meridian can only be obtained precisely by astronomical observation. The difference between true bearings at A and B is the convergence of the meridians to a point, i.e. the north pole. For small surveys the discrepancy is small and can be neglected but where necessary a correction may be computed and applied. 3.22 Magnetic north There is no fixed point and thus the meridian is unstable and sub- jected to a number of variations (Fig. 3.4), viz.: (a) Secular variation — the annual change in the magnetic declina- tion or angle between magnetic and true north. At present the magnetic meridian in Britain is to the west of true north but moving towards it at the approximate rate of lOmin per annum. (Values of declination and 116 SURVEYING PROBLEMS AND SOLUTIONS the annual change are shown on certain O.S. sheets.) (b) Diurnal variation — a daily sinusoidal oscillation effect, with the mean value at approximately 10a.m. and 6-7 p.m., and maxima and minima at approximately 8a.m. and 1p.m. (c) Irregular variation — periodic magnetic fluctuations thought to be related to sun spots. Secular variation 13001 25* 20 15 10 5 5 10 15* West East Diurnal variation East 0.00 24.00 Hours G.M.T. Fig. 3.4 Approximate secular and diurnal variations in magnetic declination in the London area (Abinger Observatory) 3.23 Grid north (see section 3.7). O.S. sheets are based on a modified Transverse Mercator projec- tion which, within narrow limits, allows: (a) Constant bearings related to a parallel grid. (b) A scale factor for conversion of ground distance to grid dis- tance solely dependent on the easterly co-ordinates of the measurement site. (See page 39). 3.24 Arbitrary north This may not be necessary for absolute reference and often the first leg of the traverse is assumed to be 0°00 '. CO-ORDINATES 117 Example 3.1 True north is 0°37'E of Grid North. Magnetic declination in June 1955 was 10°27' W. If the annual variation was 10' per annum towards North and the grid bearing of line AB 082°32' , what will be the magnetic bearing of line AB in January 1966? G.N. MM'66 MK'55 »• Fig. 3.5 Grid bearing Correction True bearing Mag declination June 1955 Mag. bearing June 1955 Variation for January 1966 -lO^xlO' Mag. bearing January 1966 082°32' -0°37 / 081°55' 10°27' 092°22' -1°45' 090°37' 3.25 Types of bearing There are two types in general use: (a) Whole circle bearings (W.C.B.), which are measured clockwise from north or 0°-360°. 118 SURVEYING PROBLEMS AND SOLUTIONS (b) Quadrant bearings (Q.B.), which are angles measured to the east or west of the N/S meridian. For comparison of bearings, see Fig. 3.6. N Case (i) Case (ii) Case (Hi) Case (iv) W 270'- W 270- Fig. 3.6 Comparison of bearings CO-ORDINATES 119 Case (i) Whole circle bearing in the first quadrant — 90° W.C.B. of AB = a° Q.B. of AB = N a?E Case (ii) 90°- 180° W.C.B. of AC = a° 2 Q.B. of AC = S 0°E = S(180-a 2 )°E Case (Hi) 180°- 270° W.C.B. of AD = a° 3 Q.B. of AD = S0°W = S(03-180)°W Case(iv) 270°- 360° W.C.B. of AE = a° 4 Q.B. of AE = N W = N (360 -a^W Example 3.2 072° = N 72° E 148° = S32°E i.e. 180-148 = 32° 196° = S 16° W i.e. 196-180 = 16° 330° = N30°W i.e. 360-330 = 30° N.B. Quadrant bearings are never from the E/W line, so that the pre- fix is always N or S. It is preferable to use whole circle bearings for most purposes, the only advantage of quadrant bearings being that they agree with the values required for trigonometrical functions 0-90° as given in many mathematical tables (see Chapter 2), e.g.: (Fig. 3.7a) sin 30° = 0-5 cos 30° = 0-8660 tan 30° = 0-5774 (Fig. 3.7b) sin 150° = sin (180 -150) = sin 30° cos 150° = -cos (180 -150) = -cos 30° tan 150°= Sinl5 ° = +sin3 ° cos 150 -cos 30 = -tan 30° 120 SURVEYING PROBLEMS AND SOLUTIONS (a) Cos 30° * E Bearing N E + + (b) -Cos 30° Bearing S E *E 210° -Cos 30* Bearing S W Bearing N W + Cos 30° + ~ 330" Fig. 3.7 (Fig. 3.7c) sin 210° = - sin (210 - 180) = -sin 30° cps210° = -cos (210- 180) = -cos 30° CO-ORDINATES 121 tan 210° = sin 210 cos 210 = +tan30 c - sin 30 -cos 30 (Fig. 3.7d) sin 330° = - sin (360 - 330) = - sin 30° cos 330° = cos (360 -330) = + cos 30° tan 330° = sin 330 - sin 30 cos 330 + cos 30 = -tan 30° 3.26 Conversion of horizontal angles into bearings. (Fig. 3.8) Fig. 3.8 Conversion of horizontal angles into bearings Forward Bearing AB = a° Back Bearing BA = a ± 180° Forward Bearing BC = a ± 180 + If the sum exceeds 360° then 360 is subtracted, i.e. Bearing BC(j8) = a ± 180 + - 360 = a + Q ± 180 This basic process may always be used but the following rules simplify the process. (1) To the forward bearing add the clockwise angle. (2) If the sum is less than 180° add 180°. If the sum is more than 180° subtract 180°. (In some cases the sum may be more than 540°, then subtract 540°) N.B. If the angles measured are anticlockwise they must be sub- tracted. 122 SURVEYING PROBLEMS AND SOLUTIONS Example 3.3 Fig. 3.9 bearing AB = 030° + angle ABC = 210° 240° - 180° bearing BC = 060° + angle BCD = 56° 116° + 180° bearing CD = 296° + angle CDE = 332° 628° - 540° bearing DE = 088° N 30° E N60°E N64°W N88°E CO-ORDINATES 123 Check bearing AB = 030° angles = 210° 56° 332° 628° -«xl80°, i.e. -3x180° - 540° bearing DE = 088° The final bearing is checked by adding the bearing of the first line to the sum of the clockwise angles, and then subtracting some multiple of 180°. Example 3.4 The clockwise angles of a closed polygon are observed to be as follows: A 223°46' B 241°17' C 257°02' D 250°21' E 242° 19' F 225°15' If the true bearings of BC and CD are 123° 14' and 200° 16' respectively, and the magnetic bearing of EF is 333°2l', calculate the magnetic declination. (N.R.C.T.) From the size of the angles it may be initially assumed that these are external to the polygon and should sum to (2n + 4)90°", i.e. l(2x6) + 4}90 = 16x90 = 1440° 223°46' 241° 17' 257°02' 250° 21' 242° 19' 225°15' Check 1440°00' To obtain the bearings, Line BC bearing 123° 14' + angle BCD 257°02' 380° 16' - 180° 124 SURVEYING PROBLEMS AND SOLUTIONS bearing CD 200° 16' (this checks with given + angle CDE 250°21' 450°37' value) - 180° bearing DE 270°37' + angle DEF 242° 19' 512°56' EF 180° bearing 332°56' + angle EFA 225° 15' 558° 11' FA 540° bearing 018°11' + angle FAB 223°46' 241°57' AB 180° bearing 061°57' + angle ABC 241°17' 303° 14' BC EF 180° bearing 123° 14' Check Magnetic bearing 333°21' True bearing EF 332°56' Magnetic declination 0°25' W 3.27 Deflection angles (Fig. 3.10) In isolated cases, deflection angles are measured and here the normal notation will be taken as: Right angle deflection— positive. Left angle deflection— negative. Taking the Example 3.3, Bearing AB Deflection right Deflection left Deflection right 030° + 30° -124° + 152° CO-ORDINATES 125 •152°(R) Fig. 3. 10 Deflection angles Bearing AB 030° + 30° Bearing BC 060° + 360° 420° -124° Bearing CD 296° + 152° 448° -360° Bearing DE 088° Check AB 030° + 30° + 152° + 212° -124° DE 088° 126 SURVEYING PROBLEMS AND SOLUTIONS Exercises 3(a) 1. Convert the following whole circle bearings into quadrant bearings: 214°30' ; 027°15' ; 287°45' ; 093°30' ; 157°30'; 311°45' ; 218°30' ; 078°45' ; 244°14' ; 278°04.' (Ans. S34°30' W; N 27°15' E; N 72°15' W; S 86°30' E; S 22°30' E; N 48°15' W; S 38°30' W; N 78°45' E; S64°14' W; N81°56' W ) 2. Convert the following quadrant bearings into whole circle bearings: N 25°30' E; S 34°15' E; S 42°45' W; N 79°30' W; S 18°15' W; N 82°45' W; S 64°14' E; S 34°30' W. (Ans. 025°30' ; 145°45' ; 222°45' ; 280°30' ; 198°15' ; 277°15' ; 115°46' ; 214°30') 3. The following clockwise angles were measured in a closed tra- verse. What is the angular closing error? 163°27'36"; 324°18'22"; 62°39' 27" ; 330° 19' 18" ; 181°09'15"; 305°58'16"; 188°02'03"; 292°53'02"; 131°12'50" (Ans. 09") 4. Measurement of the interior anticlockwise angles of a closed tra- verse ABC D E have been made with a vernier theodolite reading to 20 seconds of arc. Adjust the measurements and compute the bearings of the sides if the bearing of the line AB is N 43° 10' 20" E. Angle EAB 135°20'40" (R.I.C.S. Ans. AB N 43°10' 20" E ABC 60°21'20" BC S 17°10'52"E BCD 142°36'20" CD S 20°12'56" W CDE 89°51'40" DE N 69°38' 36" W DEA 111°50'40" EA N01°29'08" W) 5. From the theodolite readings given below, determine the angles of a traverse ABCDE. Having obtained the angles, correct them to the nearest 10 seconds of arc and then determine the bearing of BC if the bearing of AB is 45° 20' 40". Back Station Theodolite Forward Readings Station Station Back Station Forward Station E A B 0°00'00" 264°49'40" A B C 264°49'40" 164° 29' 10" B C D 164° 29' 10" 43°58'30" C D E 43°58' 30" 314° 18 '20" D E A 314°18' 20" 179°59' 10" (R.I.C.S. Ans. 125°00' 20") CO-ORDINATES 127 3.3 Rectangular Co-ordinates A point may be fixed in a plane by linear values measured parallel to the normal xy axes. The x values are known as Departures or Eastings whilst the y values are known as Latitudes or Northings. The following sign convention is used: Direction East + x West -x North +y South _ y + departure - departure + latitude - latitude + Easting (+E) -Easting (-E) + Northing (+N) -Northing (-N) N.W. quadrant (270*-360*) N Latitude N.E. quadrant (0*-90') 270* W-+ Departure (eastings) *90*E (-2,-4) S.W. quadrant (180'- 270*) (+4,-2) S.E. quadrant OC-ieO*) Fig. 3.11 Rectangular co-ordinates N.B. 0°-90° -+ NE i.e. + N +E or + lat + dep. 90°- 180° — S E i.e. -N +E -lat + dep. 180°- 270° -» S W i.e. -N -E -lat -dep. 270°- 360° -» NW i.e. + N-E + lat -dep. This gives a mathematical basis for the determination of a point with no need for graphical representation and is more satisfactory for the following reasons: (1) Each station can be plotted independently. (2) In plotting, the point is not dependent on any angular measur- ing device. (3) Distances and bearings between points can be computed. 128 SURVEYING PROBLEMS AND SOLUTIONS Rectangular co-ordinates are sub-divided into: (1) Partial Co-ordinates, which relate to a line. (2) Total Co-ordinates, which relate to a point. 3.31 Partial co-ordinates, AE, AN (Fig. 3.12) These relate one end of a line to the other end. They represent the distance travelled East (+)/West (-) and North (+)/South (-) for a single line or join between any two points. Fig. 3.12 Partial co-ordinates Given a line of bearing d and length s, Partial departure = AE i.e. difference in Eastings A E AB = s sin Q Partial latitude = AN i.e. difference in Northings AN^g = scosd (3.1) (3.2) N.B. always compute in bearings not angles and preferably quadrant bearings. 3. 32 Total co-ordinates (Fig. 3. 13) These relate any point to the axes of the co-ordinate system used. The following notation is used: Total Easting of A = E A " Northing of A = N^ Total Easting of B = E A + A E AB « Northing of B = N A + AN AB Total Easting of C = E B + A E BC = E A + A E AB + A E BC " Northing of C = N B + &N BC = N^ + AN^ + AN BC CO-ORDINATES 129 Fig. 3.13 Total co-ordinates Thus in general terms Total Easting of any point = E^+SAE (3.3) = Total easting of the first point + the sum of the partial eastings up to that point. Total Northing of any point = N^ + 2 AN (3.4) = Total northing of the first point + the sum of the partial northings up to that point. N.B. If a traverse is closed polygonally then SAE = (3.5) SAN = (3.6) i.e. the sum of the partial co-ordinates should equal zero. Example 3.5 Given: (Fig. 3.14) AB 045° 100 m BC 120° 150 m CD 210° 100 m Total co-ordinates of A E 50 m N 40 m Line AB 045° = N 45° E 100 m Partial departure AE^ = 100 sin 45° = 100 x 0-707 = + 70*7 m Total departure (E^) A = + 50*0 m Total departure (E fl ) B= + 120-7 m 130 SURVEYING PROBLEMS AND SOLUTIONS -50* (200-6.-50-9) Fig. 3.14 Partial latitude A N^ = 100 cos 45°= 100x0*707 = + 70-7 m Total latitude (N^) A = + 40*0 m Total latitude (N 5 ) B= + 110*7 m Line BC 120° = S 60° E 150 m Partial departure AE BC = 150 sin 60° = 150 x 0*866 = + 129*9 m Total departure (E B ) B = + 120*7 m Total departure (E C )C = + 250*6 m Partial latitude AN^ C = 150 cos 60° = 150 x 0*5 = - 75*0 m Total latitude (N B )fi = +110*7 m Total latitude (N c ) C = + 35*7 m Line CD 210°= S30°W 100 m Partial departure AE CD = 100 sin 30° = 100 x 0*5 = - 50*0 m Total departure (E^) C = + 250*6 m Total departure (E^) D = + 200*6 m Partial latitude AN CD = 100 cos 30° = 100 x 0*866 = - 86*6m Total latitude (N c ) C = + 35*7 m Total latitude (N^) D = - 50*9 m Check E D = E A + &E AB + AE BC + AE CZ? = 50*0 + 70*7 + 129*9 - 50*0 = + 200*6 m CO-ORDINATES 131 N^= K A + Mi AB + AN SC + ANqd = 40-0 + 70-7 - 75-0 - 86'6 = - 50-9 m Exercises 3(b) (Plotting) 6. Plot the following traverse to a scale of 1 in = 100 links, and thereafter obtain the length and bearing of the line AB and the area in square yards of the enclosed figure. N 21° W 120 links from A N 28° E 100 links N60°E 117 links N 32° E 105 links S 15° E 200 links S 40° W 75 links to B (Ans. From scaling N62°45'E 340 links; approx. area 3906 sq yd,) 7. The following table shows angles and distances measured in a theodolite traverse from a line AB bearing due South and of horizontal length 110 ft. Angle Angle value Inclination Inclined distance (ft) ABC 192°00' +15° BC 150 BCD 92°15' 0° CD 200 CDE 93°30' -13° DE 230 DEF 170°30' 0° EF 150 Compute the whole circle bearing of each line, plot the survey to a scale of 1 in. = 100 ft and measure the horizontal length and bear- ing of the closing line. (M.Q.B./M. Ans. 260 ft; 076°30') 8. The following notes reter to an underground traverse made from the mouth, A, of a surface drift. Line Bearing Distance (links) AB 038° 325 dipping at 1 in 2-4 BC 111° 208 level CD 006° 363 level DE 308° 234 rising at 1 in 3-2 Plot the survey to a scale of 1 chain to 1 inch. Taking A as the origin, measure from your plan, the co-ordinates of E. What is the difference in level between A and E to the nearest foot? (M.Q.B./UM Ans. E, E 233 links N 688 links; diff. in level AE 78 ft) 132 SURVEYING PROBLEMS AND SOLUTIONS 9. Plot the following notes of an underground traverse to a scale of lin ** 100 ft. Line Bearing Distance AB N 28° W 354 ft dipping at 1 in 7 BC N 83° W 133 ft level CD S 83° W 253ft level DE N 8°E 219 ft rising at 1 in 4 EF S 89° E 100 ft level Points A,B,C and D are in workings of a lower seam and points £ and F are in the upper seam, DE being a cross measure drift be- tween the two seams. It is proposed to drive a drift from A to F. Find the bearing, length, and gradient of this drift. (M.Q.B./UM Ans. N 40° W; 655 ft; +1 in 212) 10. The co-ordinates in feet, relative to a common point of origin A, are as follows: Departure Latitude A B 275 E 237 N C 552 E 230 N D 360 E 174 S Plot the figure A BCD to a scale of 1 inch to 100 ft and from the co-ordinates calculate the bearing and distance of the line AC. (M.Q.B./UM Ans. N 67°24' E; 598ft) 11. An area in the form of a triangle ABC has been defined by the co-ordinates of the points A B and C in relation to the origin 0, as follows: A South 2460 ft East 3410 ft B North 2280 ft East 4600 ft C North 1210 ft East 1210 ft Plot the positions of the points to a scale of 1 in. to 1000 ft, and find the area, in acres, enclosed by the lines joining AB, BC and CA. (M.Q.B./M Ans. 169-826 acres) 12. There is reason to suspect a gross angular error in a five-legged closed traverse in which the recorded information was as follows: Interior angles: A 110°; B 150°; C 70°; D 110°; E 110° Sides: AB 180 ft; BC 420 ft; CD 350 ft; DE 410 ft; EA 245 ft CO-ORDINATES 133 Plot the traverse to a scale of 100 ft to lin. and locate the gross angular error*, stating its amount. (L.U./E) 13. A rough compass traverse of a closed figure led to the following field record: Line Length Bearing AB 422 57° EC 405 316° CD 348 284° DE 489 207° EA 514 109° Plot the figure (scale lin = 50ft) and adjust it to close using a graphical method. Letter your plan and add a north point (magnetic declination 10° W). (L.U./E) 3.4 Computation Processes As tables of trigonometrical functions are generally tabulated only in terms of angles 0°-90°, it is convenient to convert the whole circle bearings into reduced or quadrant bearings. The signs of the partial co-ordinates are then related to the sym- bols of the quadrant bearings, Fig. 3.15. E + 1 r^ N > Departures W - J ::) Latitudes Alternatively, the whole circle bearings are used and the sign of the value of the partial co-ordinate is derived from the sign of the trigonometrical function. W + N W + - N E + + + sw SE - + — s Fig. 3. 15 The process may be either : (i) by logarithms or (ii) by machine (using natural trigonometrical functions). See Chapter 6 on location of errors. 134 SURVEYING PROBLEMS AND SOLUTIONS 3.41 Computation by logarithms Let AB = 243° 27' 423-62 m (.4 2063-16 m E 5138 -42 m N) (243° 27' = S63°27' W) Logs E A 2063-16 m partial departure (AE) 2-578 579 sin bearing AB 1-951 602 \ distance 2-626977 1\ cos bearing 1 -650 287 ) partial latitude (&N AB ) 2-277 264 N.B. The log distance is written down once only, being added to the log sin bearing above and the log cos bearing below, to give the partial departure and latitude respectively. It is often considered good computing practice to separate the log figures for convenience of adding though the use of squared paper would obviate this. 3.42 Computation by machine t^AB -378-95 E B 1684-21 m "a 5138-42 AN AB -189-35 N B 4949-07 m E A 2063-16 partial departure AR._ AB.o -378-95 E B 1684-21 m sin bearing 0-894 545 J 23-62 J \ 0-446 979 1 distance A cos bearing "a 5138-42 partial latitude AN > W AB 189-35 ^MB * "b 4949-07 in Using a normal digital machine, the distance (being common) is set once in the machine and then separately multiplied by the appropri- ate trigonometrical function. In the case of the twin-banked Brunsviga, the processes are simul- taneous. N.B. For both natural and logarithmic trigonometrical functions the following tables are recommended: Degrees only 4 figure tables Degrees and minutes 5 figure tables Degrees, minutes and seconds 6 figure tables Degrees, minutes, seconds and decimals of seconds 7 figure tables CO-ORDINATES 135 3.43 Tabulation process (Fig. 3.16) Nottingham Regional College of Technology Traverse computation sheet N* 1 Traverse A to Z Compiled by _^±y_- (Red) Computed by .^/i.jGreen) Date _?j_r„/ii.7 : . Checked by _!??•„. 4E = length x sin B'g 4N= length x cos B'g Line Bearing Length sin/cos S£r AE AH E N Ftoint AB 243* 27'00" S63*27'W 423-62 s 0-894 545 C0446 979 -378 95 -189-35 2063-16 5138-42 A 1684-21 494907 B BC 042* 32' 00" N42*32' E 221-38 SO676 019 cO-736 884 H49-66 +163-13 1833 87 5112-20 C Check £ -229-29 2063-16 -26-22 5138-42 1833-87 5112-20 Fig. 3. 16 Tabulated computation Example 3.6 Calculate the total co-ordinates, in feet, of a point B if the bearing of AB is 119° 45' and the distance is 850 links on a slope of 15° from the horizontal. The co-ordinates of A relative to a local origin are N 5356*7 ft E 264-5 ft. (M.Q.B./UM) To find horizontal length (Fig. 3.17) Fig. 3. 17 AB, = AB cos 15° links but 100 links = 66 ft; therefore to convert links to feet the length must be multiplied by K = 0*66, i.e. AB, = K.AB cos 15° = 0-66 x 850 x cos 15° By logs, 0-66 1-81954 850 2-92942 cos 15° T-984 94 AB, 2-73390 136 SURVEYING PROBLEMS AND SOLUTIONS To find partial co-ordinates of line AB (Fig. 3.18) Fig. 3. 18 119° 45' = S60°15' E By logs, partial departure 2*672 52 A£ 'AB sin 60° 15' AB, cos 60° 15' 1-938 62 2-73390 (see above) 1-695 67 partial latitude 2-429 57 AN + 264-5 + 470-4 E B +734-9 5356-7 -268-9 'AB N D 5087-8 Co-ordinates of B, N 5087-8 ft E 734-9 ft 3.44 To obtain the bearing and distance between two points given their co-ordinates (Fig. 3. 19) Let the co-ordinates of A and B be E^N^ and E s N fl respectively. F _ F Then tan bearing (0) = —2 ± N - N ™B n A AE •AB AN AB N.B. For convenience this is frequently written : Bearing AB = tan-'AE/AN (the sign of the differences will indicate the quadrant bearing). Length AB = >/(AE* + AN* ) (3.7) (3.8) (3.9) (3.10) CO-ORDINATES 137 N.B. This is not a very good solution for computation purposes and the trigonometrical solution below is preferred. or AB = AB = cos bearing (0) AN^g sec# AE AB (3.11) (3.12) (3.13) sin bearing (0) = AE^cosectf (3.14) If both of these determinations are used, their agreement provides a check on the determination of 0, but no check on the subtraction of the Eastings or Northings. AZab N„ 1 'B *"ab , ^a . ^/* >B N/, • A E A E ■B Fig. 3.19 To find the length and bearing between two points Example 3.7 A B E 632-16 m 925-48 m N 949-88 m 421-74 m AE + 293-32 m AN - 528-14 Bearing AB = tan"' + 293 ' 32 (E) -528-14 (S) 138 SURVEYING PROBLEMS AND SOLUTIONS By logs, 293-32 2-467 34 528-14 2-72275 tan (0) 1-744 59 — Length AB = AN sec 6 = 528-14 sec 29° 03' » S 29° 03' E i.e. 150^57' or A E cosec 6 293-32 cosec 29° 03' By logs, or 528-14 sec 29° 03' 293-32 cosec 29° 03' 2-72275 0-058 39 2-781 14 2-46734 0-31375 2-78109 -* 604-14 m -> 604-07 m The first solution is better as AN > AE, but a more compatible solution is obtained if the bearing is more accurately determined, using 7 figure logs, or AE 2-4673417 AN 2-7227491 tan ((9) 9-7445926 Bearing (0) = 29°02'50" AN 2-7227491 sec# 10-058 3786 2-7811277 AE 2-467 3417 cosec 10-3137836 - 604-13 m 2-7811253 604- 12 m Example 3.8 The following horizontal angle readings were recorded during a counter-clockwise traverse ABCD. If the line AD is taken as an arbitrary meridian, find the quadrantal bearings of the remaining lines. Find also the latitudes and departures of the line CD whose length is 893-6 m. COORDINATES 139 Station at Sight Vernier A Vernier B A D 241° 36' 20" 061° 36' 40" A B 038° 54' 00" 218° 53' 40" B A 329° 28' 00" 149° 28' 20" B C 028° 29' 00" 208° 29' 00" C B 106° 58' 20" 286° 58' 40" C D 224° 20' 20" 044° 20' 20" D C 026° 58' 00" 206° 58' 40" D A 053° 18' 40" 233° 18' 00" Ans. Mean values A D Angle 241° 36' 30" B 038° 53' 50" DAB 157° 17' 20" B A 329° 28' 10" C 028° 29' 00" ABC 59°00'50" C B 106° 58' 30" D 224° 20' 20" BCD 117° 21' 50" D C 026° 58' 20" A 053° 18' 20" CDA 26°20'00" 360° 00 '00" Bearing AD = 0°00' Angle DAB = 157° 17' 20" Bearing AB = 157° 17' 20" (S 22° 42' 40" E) Angle ABC = 59° 00' 50" 216° 18' 10" -180° Bearing BC 036° 18' 10" (N 36° 18' 10" E) Angle BCD 117° 21' 50" 153° 40' 00" + 180° Bearing CD 333°40'00 y/ (N 26°20'00" W) Angle CD/ I 26° 20' 00" 360° 00' 00" Co-ordinates CD 893-6 m (N26°20'W) AE = 893-6 sin 26° 20' = -396-39 m AN = 893-6 cos 26° 20' = +800-87m 140 SURVEYING PROBLEMS AND SOLUTIONS Ans. AB = S 22° 42' 40" E BC = N36°18'10"E CD = N26°20'00" E Co-ordinates of line CD AE = -396 -4m AN = +800*9 m Example 3.9 In order to continue a base line AC to G, beyond a building which obstructed the sight, it was necessary to make a trav- erse round the building as follows, the angles being treated as deflec- tion angles when traversing in the direction ABCDEFG. ACD = 92° 24' to the left CDE = 90° 21' to the right CD = 56-2 ft DEF = 89° 43' to the right DE = 123-5 ft Calculate EF for F to be on AC produced and find EFG and CF. (L.U.) IG i I 89°43' £ ^_J $__».-- r i i i I i i i i I I i X. 1- i i I I I I I I i I I I 90°21y 92°24', 1 -o- Fig. 3.20 Assuming the bearing AC = 0°00' CO-ORDINATES 141 i.e. Bearing AC = 360° 00' -angle ACD (left) 92° 24' Bearing CD 267° 36' i.e. S 87° 36' W + angle CDE (right) 90 o 2l' Bearing DE 357° 57' i.e. N 02° 03' W + angle DBF (right) 89° 43' 447° 40' -360° 00' Bearing EF 087° 40' i.e. N 87° 40' E Thus, to obtain the bearing of FG = bearing AC, the deflection angle EFG = 87° 40' left. Check on deflection angles + 90° 21' 92° 24' 89° 43' 87° 40' 180° 04' 180° 04' To obtain the co-ordinates of E Line Distance (ft) Bearing sin Bearing cos Bearing Ae An AC o°oo' CD 56-2 S 87°36'w 0-999 12 0-04188 -56-15 - 2«35 DE 123-5 N02°03'w 0-035 77 0-999 36 - 4-42 + 123*42 + 123.42 - 2.35 E F - 60.57 N^ +121-07 Thus F must be +60*57 ft east of E. The line EF has a bearing 087° 40' Length EF = AE ^ = 60 ' 57 , sin bearing sin 87° 40' = 60-62 ft To find the co-ordinates of F, &N EF = 60-62 cos 87° 40' = + 2-47 N £ = +121-07 N/r = +123-54 142 SURVEYING PROBLEMS AND SOLUTIONS F is 123*54 ft above C on the bearing due N. .-. CF = 123-54 ft. Ans. EF = 60-6 ft EFG = 87° 40' deflection left CF = 123-5 ft Example 3.10, The co-ordinates (metres) of the base line stations A and B are A 26 543-36 E 35432-31 N B 26895-48 E 35983-37 N The following clockwise angles were measured as part of a closed traverse: ABCDEA ABC 183° 21' BCD 86° 45' CDE 329° 17' DEA 354° 36' EAB 306° 06' Determine the adjusted quadrant bearings of each of the lines rela- tive to the meridian on which the co-ordinates were based. A 26 543-36 m 35 432-31 m B 26895-48 m 35 983-37 m AE 352-12 m AN 551-06 m tan bearing AB = 352-12 551-06 bearing AB = 032° 34' corr. Corrected Angle 2 Angles 183° 21' -01' 183° 20' 86° 45' -01' 86° 44' 329° 17' -01' 329° 16' 354° 36' -01' 354° 35' 306° 06' -01' 306° 05' 1260° 05' 1260° 00' £ Angles should equal (2n + 4) 90 i.e. (2x5 + 4)90 = 1260< .*. error is 05' distributed as 01' per angle. CO-ORDINATES 143 Calculation of bearings Bearing AB 032° 34' - > N 32° 34' F Angle ABC 183° 20' 215° 54' - 180° Bearing BC Angle BCD 035° 54' - > N ^5° 54' F 86° 44' 122° 38' + 180° Bearing CD 302° 38' * N57°22'W Angle CDE 329° 16' 631° 54' -540° Bearing DE 091° 54' - i ^ ftR°nfi' f Angle DEF 354° 35' 446° 29' -180° Bearing EA 266° 29' - » S 86°29'W Angle EAB 306° 05' 572° 34' -540° Bearing AB 032° 34' Check Example 3.11 A disused colliery shaft C, situated in a flooded area, is surrounded by a circular wall and observations are taken from two points A and 6 of which the co-ordinates, in feet, relative to a local origin, are as follows: Station Eastings Northings A 3608-1 915-1 B 957-6 1808-8 C is approximately N.W. of A. Angles measured at A to the tangential points 1 and 2 of the walls are BAC, = 25° 55' and BAC Z = 26° 35'. Angles measured at B to the tangential points 3 and 4 of the wall are C 3 BA = 40° 29' and C A BA = 39° 31'. Determine the co-ordinates of the centre of the shaft in feet rela- tive to the origin, to one place of decimals and calculate the diameter 144 SURVEYING PROBLEMS AND SOLUTIONS of the circle formed by the outside of the wall. A B AE 'AB Fig. 3.21 E 3608-1 957-6 _ 2650-5 N 915-1 1808-8 AN„ D + 893-7 VU3 In Fig. 3.21, Bearing of AB = tan -iAE _ AN = tan" -2650-5 + 893-7 (M.Q.B./S) = N71°22'W i.e. 288° 38' Length AB = AN sec bearing or AE cosec bearing = 893-7 sec71° 22' or 2650-5 cosec 71° 22' = 2797>1 2797-1 In triangle ABC Angle A = I {25° 55'+ 26° 35'} Angle fi = I { 40° 29' +39° 31'} Angle C = 180° - (26° 15' + 40° 00') = 26° 15' = 40° 00' = 113° 45' 180° 00' CO-ORDINATES 145 By the sine rule, BC = AB sin A cosec C = 2797-1 sin 26° 15' cosec 113° 45' = 1351-6 ft AC = BC sin B cosec A = 1351-6 sin 40° 00' cosec 26° 15' = 1964-3ft Bearing AB 288° 38' + Angle BAC Bearing AC Bearing BA - Angle CBA Bearing BC To find co-ordinates of C Line BC 068° 38' i.e. N68°38'E 1351-6 ft 26° 15' 314° 53' 108° 40° 38' 00' 068° 38' AE^ = 1351-6 sin 68° 38' = +1258-7 E c = E s + ^E BC = 957-6 + 1258-7 = +2216-3 AN 5C = 1351-6 cos 68° 38' = + 492-4 N c = N B + AN SC = 1808-8 + 492-4 = +2301-2 Check Line AC 314° 53' i.e. N45°07'W 1964-3 ft AE^ = 1964-3 sin 45° 07' = -1391-8 E C = E A + & E AC = 3608-1 - 1391-8 = 2216-3 AN^ C = 1964-3 cos 45° 07' = +1386-1 N c = N A + AK AC = 915-1 + 1386-1 = 2301-2 To find the diameter of the wall . Referring to Fig. 3. 21 a = |(40 o 29'-39°31') = 0°29' R = BCsinO°29' cz BC x 0°29'(rad) = 1351-6 x 29 x 60 = 11.40 ft 206265 146 SURVEYING PROBLEMS AND SOLUTIONS Check p = 1(26° 35' -25° 55') = 0°20' R = AC x 0°20'(rad) 1964-3 x 20 x 60 = n^ft 206 265 Diameter of wall 22-8 ft 3.5 To Find the Co-ordinates of the Intersection of Two Lines 3.51 Given their bearings from two known co-ordinate stations As an alternative to solving the triangle and then computing the co-ordinates the following process may be applied: Given A (E A N A ) B (E 5 N fl ) bearings a and jQ C (E C N C ) Fig. 3. 22 From Fig. 3.22, (3.15) (3.16) E c = E A + (N^-N^tana = E A + AN^ tana = E B + (N c -N B )tan/8 = E B + AN sc tan0 .-. N c (tana -tanjS) = E B - E A + N^ tana - N B tan0 Then the total northing of C N = E g - E^ + H A tana - N B tanft (3 1?) c ~ tana - tanjS To obtain the partial co-ordinates from equation (3.17) CO-ORDINATES 147 Partial Northing Mi AC = N c - N A 1,C N - N = Eg ~ Ea + Na tan a ~ Ng ta " ^ - N^ c A tana - tan/8 E 5 - E^ + N^tan a - N^tanft - N^ tana + N A tanft tana - tanft _ (Es-E.0 -(N B -K4) tanft tan a - tan ft -/IS tana - tan/3 Then AN^ = A E^, - A IWa. (3 lg) Then W BC = "T f. (3-19) Similarly, from equation (3.17), AU BC = N^-N* = Eg-E^ + N^tana-Nstanft tana - tanft & E AB - &N AB tan a tana - tan/8 The following alternative process may be used: N c = N4 +(E c -E x )cota = N^ + AE^ cot a (3.20) = N B + (E c -E 5 )cotft = N B + AE 5C cotft (3.21) As before, the total and partial co-ordinates are given as: E = Nb ~ Na + Ea cot a ~ E B cot ft n 2? . C cot a - cot ft ^ ' and AE^ m *B ~ AE^cotft (3.23) (3.24) cot a - cot ft A E = AN4 g - AE^ cot a cot a - cot ft N.B. Theoretically, if Scot > Stan, then it is preferable to use the cot values, though in practice only one form would be used. Example 3.12 Let the co-ordinates be A = E4, N6 8 = E13, N4 the bearings be a = 060° ft = 330° tana = 1-7321 cot a = 0-5774 tan ft = -0-5774 cot ft = -1-7321 Using the tan values ; from equation (3.17) N = (13 - 4) + (6 x 1-732 1) + (4 x 0-577 4) = g . from equation (3. 15) c = _ - - -• ' - = y-397 1-7321 + 0-5774 E c = 4 + (9-397-6) x 1*7321 = 9-884 148 SURVEYING PROBLEMS AND SOLUTIONS or equation (3. 16) E c = 13 + (9-397 -4) x -0-5774= 9-884 Using the cot values, from equation (3.22), E„ - 4 - 6 + 4 x 0-5774+ 13 x 1-7 321 Q .oo d C 0-577 4+1-7321 = ^ From equation (3.20), N c = 6 + (9-884 - 4) x 0-577 4 = 9-397 or equation (3.21) N c = 4 + (9-884-13) x -1-7321 = 9-397 If the formulae using partial values are employed the individual equation computation becomes simplified. Using the previous values (Ex. 3.5), from equation (3.18) W Ae . (13 -?-<1-?JL: 0,S774 - + 3-397 AC 1-7321 + 0-577 4 Then N c = N^ + AN^ = 6 + 3-397 = 9-397 When this value is known, equation (3.15) may be used as before. From equation (3.23); (4-6) - (13-4) x -1-7321 AEUc = E 0-577 4 + 1-7321 c = e a + ^AC = 4 + 5-884 = 9-884 = +5-884 When this value is known, equation (3.20) may be used as before. The above process is prefered and this can now be given in a tabulated form. Example 3.13 (1) AN _ &E AB - AN^tan/8 'AC — — — — — — — — ^— — — — I— tan a - tan /S (2) AE^ = tOA AC tana (3) AN 5e = AE^ B - b& A B tana tana - tan^S (4) AE BC = AN flC tan/3 CO-ORDINATES 149 Igloo Oriented diagram E base Stations E Bearings N _E_B_ase_ ^AX + 13 486-85 m a 278° 13' 57" + 10 327-36 m _Iglpo__(B) + 12 759-21 m 182° 27' 44" + 13 142-72 m tana -6-911745 2 &AB -727-64 tanjS + 0-0430004 tan a - tan + 2815-36 AN^tan/8 + 121-06 -848-70 -J- -6-9547456" * x tana — &&AC -843-44 + 122-03 W Base E c 12 643-41 m 10 449-39 m Check & E AB -727-64 Afy^tana -19459-05 tan a - tan /8 + 18 731-41 + -6-9547456" « — x tan/3 — ' & BC 115-81 2693-33 W Base E c + 12643-40 m 10 449-39 m AN l AB AN, l BC 3.52 Given the length and bearing of a line AB and all the angles A,B and C, Fig 3. 23 Given: (a) Length and Bearing of AB, (b) Angles A,B and C. E c - E^ = b cos(A +6) = b(cosA cos 6 - sin A sin#) c sin B cos A cos^ - c sinB sin 4 sin 5 but E« - E\d = c cos Q sinC j4B sin bearing^ SURVEYING PROBLEMS AND SOLUTIONS C Fig. 3. 23 N, and Then E,7 — E. = . - N4 = c sintf = AB cos bearing^ C = 180 - 04 + B) sinC = sin A cosB + cos A sin 6 (E B - E4) sinB cos 4 - (N B - N^) sing sin ,4 , sin A cos B + cos A sin S E4 sin A cos g + E^ cos A sin B + Eg sin B cos 4 - E^ sin g cos 4 - N B sin B sinA + Ha sin g sin A sin 4 cos g + cos A sin B E A cotB + E B cot 4 + (N.4 - N B ) cot A + cotB E.4 cotg + E fl cot /I - AN^s Similarly, cot 4 + cotB N^ = N,4 cotB + N B cot ,4 + &E AB (3.25) (3.26) cot A + cotB Check E^cotB - 1) + E fl (cot4 + 1) + N^cot B + 1) + + N^cotA - 1) - (E c + N c )(cot4 + cot B) = (3.27) Using the values of Example 3.13, Bearing AB = tan-' - 727-64/+ 2815-36 = N 14° 29' 28" W = 345° 30' 32" Bearing AC = 278° 13' 57" .'• Angle A = 67° 16' 35" Bearing BC = 182° 27' 44" Bearing BA = 165° 30' 32" ••• Angle B = 16° 57' 12" Bearing CB = 02° 27' 44" Bearing CA = 098° 13' 57" .'• Angle C = 95° 46' 13" cot A = 0-41879 From equation (3.25), CO-ORDINATES 151 Check 1 = 180° 00' 00" cotB = 3*280 40 cotC = -0-10105 P *L A cot B + E B cot A - /SNab & c = cot .4 + cotB (13486-85 x 3-2804) + (12759-21 x 0-41879) - 2815-36 0-418 79 + 3-28040 44 242-26 + 5343-43 - 2815-36 E c = 12643-40 From equation (3.26), 3-699 19 N^, = N,4 cot B + N B cot A + A E AB cot A + cotB (10 327-36x3-2804) + (13142-72x0-41879) - 727-64 3-699 19 33877-87 + 5504-04 - 727-64 3-699 19 N c = 10 449-39 Check (equation 3.27) E/cotB-l) = 13486-85 x 2-28040 = 30755-41 E^cot^ + l) = 12759-21 x 1-41879 = 18102-64 N^(cotB + l) = 10 327-36 x 4-280 40 = 44 205-23 N B (cotA - 1) = 13142-72 x -0-58121 = - 7638-68 = 85 424-60 (E<? + N c Xcot^ + cotS) = (12643-40 + 10449 -39)(3-69919) = 85424-62 Example 3.14 Given the co-ordinates of four stations, B C A tL zoO-00 N 100-00 B E 320-70 N 170-70 C E 520-70 N 170-70 D E 652-45 S 263*12 Fig. 3.24 to find the co-ordirtates of the intersection of the lines AC and BD. 152 SURVEYING PROBLEMS AND SOLUTIONS Method 1 AE tan bearing BD = -r__ AN 652-45 - 320-70 331-75 -263-12 - 170-70 -433-82 bearingSD = S37°24'E = 142° 36' = bearingfiX 250-0 - 520-7 -270-7 tan bearing CA = 100-0 - 170-7 -70-7 bearing CA = S75°22'W = 255° 22' = bearing CX In triangle BCX BC = 520-7 - 320-7 = 200 (No difference in latitude, therefore due E) Bearing BC = 090° BX = 142° 36' .". Angle XBC = 52° 36' Bearing CB = 270° CX = 255° 22' Angle BCX = 14° 38' Length BX = BC sinBCX = 200 sin 14° 38' cosec (52° 36'+ 14° 38') sin BXC logBX = 1-73875 To find co-ordinates of X. (Length BX known. Bearing S37°24'E) Logs ; 1-522 21 > + 33-28 (AE) E4 + 320-7 sin bearing 1*78346 AE 33-28 length 1-73875 E* 353-98 cos bearing 1-90005 1-63880 > - 43-53 (AN) N* + 170-70 AN - 43-53 Ans. X = E 353-98 N 127-17 N* + 127-17 Method 2 From the previous method, tan bearing BD (/S) = Zj&k = -°' 76472 tan bearing CA (a) = ~ ' ' = 3-82885 -70-7 CO-ORDINATES 153 Using equation (3.18), AN = AE CB - AN gg tanj8 tan a - tan )S AE^ = AN cx tana E N B 320-70 170-70 C 520-70 170-70 AE CB - 200-00 AN CB 0-0 tana - tan0 = 3-828 85 + 0-76472 = 4-59357 AN ~ - ?Hf - ° - - 43 - 538 N c = 170-70 N x = 127-16 AE CX = -43-538 tana = -43-538x3-82885 = -166-70 E c = 520-70 E^ = 354-00 Method 3 By normal co-ordinate geometry, the equation of line AC = ^ ~ y i = y z ~ ^i i.e. y-100 = 170-7 - 100 = 70j7 = x-250 520-7-250 270-7 uzolz y _100 = 0-261 2 (x - 250) (1) Similarly, the equation of line BD = * ~ 170 ' 7 = -263-12 - 170-7 x - 320-7 652-45 - 320-7 = ~ 433 ' 82 = -1-3076 331-75 i-e. y- 170-7 = - 1-3076 (x - 320-7) (2) Subtracting (1) from (2), 70-7 = 1-5688 x - (250 x 0-2612) - (320-7 x 1-3076) = 1-568 8 x _ 65-3 - 419-343 2 x = 555 ' 3432 = 354-00 1-5688 154 SURVEYING PROBLEMS AND SOLUTIONS Substituting in equation (1), y = 0-261 2 (x- 250) + 100 = 0-2612(354-250) + 100 = (0-2612 xl04) + 100 = 127-16 Ans. X = E 354-0 N 127-16 N.B. All these methods are mathematically sound but the first has the advantages that (1) no formulae are required beyond the solution of triangles, (2) additional information is derived which might be re- quired in setting-out processes. Example 3.15. Equalisation of a boundary line. The following sur- vey notes refer to a boundary traverse and stations A and E are situ- ated on the boundary. Horizontal length (ft) 253-2 426-4 543-8 1260-2 It is proposed to replace the boundary ABCDE by a boundary AXE where AX is a straight line and X is situated on the line DE. Calculate the distance EX which will give equalisation of areas on each side of the new boundary. (M.Q.B./S) Computation of co-ordinates with A as the origin, Fig. 3.25 Line AB N 83° 14' E 253-2 ft Line Bearing AB N 83° 14' E BC S 46°30'E CD N 36°13'E DE S 23°54'E Logs E, 0-0 AE 2-400 42 9-99696 + 251-44 sin# E B + 251-44 length 2-40346 cos 6 9-071 24 N A 0-0 AN 1-47460 + 29-83 No + 29-83 Line BC S46°30'E 426-4 E B + 251-44 AE 2-490 38 -» + 309-30 sin0 9-86056 CO-ORDINATES 155 length 2-62982 cos 6 9-83781 AN 2-46763 N 5 + 29-83 - 293-51 N c - 263-68 Line CD N 36° 13' E 543-8 E c + 560-74 AE 2-50691 -> + 321-30 sin0 9-77147 E D + 882-04 length 2-73544 cos# 9-90676 N c - 263-68 AN 2-642 20 _+ 438-74 H D + 175-06 Fig. 3.25 -200 -400 -600 ■800- -1000 200 400 800 800 1000 1200 1400 By construction, Join BD. Draw line parallel to BD through C to cut ED at C, . Area of triangle BDC X = area of triangle BDC (triangles on same base and between same parallels). Join C X A. Draw line parallel to C,/4 through B to cut ED at 8,. Area of triangle AB^C = area of triangle ABC^. •'• Line AB X (X) equalises the irregular boundary in such a way that triangle ABP + triangle QDB^ = triangle PQC Length EX = 977.84 ft (calc.) = 978 ft (scaled). 156 SURVEYING PROBLEMS AND SOLUTIONS Line DE S 23° 54' E 1260-2 E D + 882-04 AE 2-70805 + 510-57 sin0 9-60761 length 3-100 44 E E + 1392-61 costf 9-96107 N^ + 175-06 AN 3-06151 - 1152-15 N^ - 977-09 Checks AE + 251-44 AN + 29-83 - 293-51 + 309-30 + 438-74 -1152-15 + 321-30 + 468-57 -1445-66 2AE + 510-57 + 1392-61 + 468-57 SAN - 977-09 Area of figure ABCDEA (see ft) (2) Chapter 11) (3) (4) (5) N E F.dep. B.dep. (3) (4) A 0-0 0-0 + 251-44 +1392-61 -1141-17 B + 29-83 + 251-44 + 560-74 0-0 + 560-74 C - 263-68 + 560-74 + 882-04 + 251-44 + 630-60 D + 175-06 + 882-04 + L392-61 + 560-74 + 831-87 E - 977-09 + 1392-61 0-0 + 882-04 - 882-04 Double areas (1) x (5) A B 167 26-8 C 166 276-0 D 145627-0 E 861832-0 + 1024185-8 - 166 276-0 2; 857 909-8 428 954-4 sq ft From co-ordinates Bearing EA = tan~ 1 - 1 ^ 92 ' 61 = N 54° 56' 44" W = 305° 03' 16" CO-ORDINATES 157 - ' A A I' Length EA = 1392-61 cosee 54° 56' 44' = 1701-2 ft (x) Bearing ED = N 23° 54' 00" W = 336° 06' 00" Angle AED = 336° 06' 00" - 305° 03' 16" = 31° 02' 44" To find length EX (a) such that the area of triangle AXE is equal to 428 954-4 sq.ft. Area of triangle AXE = ^ax sin AED - 2 area triangle AXE x sin AED 2 x 428 954-4 1701-2 sin31°02' 44" = 977-84 ft (length EX). Exercises 3(c) (Boundaries) 14. The undernoted bearings and measurements define an irregular boundary line on a mine plan between two points A and B, the latter being a point on a straight line XY , bearing from South to North. Plot the bearings and measurements to a scale of 1/2 in. = 100 ft, and thereafter lay down a straight line from A to a point on XBY so that the areas to the North and South respectively of that line will be equal. From A N 63° 30' W 185 ft S45°00'W 245 ft S80°45'W 175 ft N 55° 15' W 250 ft S 60° 30' W 300 ft to B Check your answer by calculation of the respective areas. (M.Q.B./M) 15. The undernoted traverse was taken along an irregular boundary between two properties: AB N32°45'E 464 ft BC N71°30'E 308 ft CD S61°15'E 528 ft DE N71°30'E 212 ft EF S40°30'E 248 ft A lies on a straight boundary fence XY which bears N 7° 30' W. 158 SURVEYING PROBLEMS AND SOLUTIONS Plot the traverse to a scale 1/2400 and thereafter set out a straight line boundary from F to a point G on the fence XAY so that the areas North and South of the line are equal. What length of fencing will be required ? How far is G from A ? (N.R.C.T. Ans. 1480 ft; AG 515 ft) 3.6 Transposition of Grid New axis Fig. 3. 26 Transposition of grid Let the line AB (Fig. 3.26) based upon an existing co-ordinate system have a bearing 6 and length s. Then AE^ = s sin (9 ^AB = s COS0 The co-ordinate system is now to be changed so that the origin of the new system is 0. The co-ordinates of the position of 'slew' or rotation A = E^ and the axes are rotated clockwise through an angle +a to give a new bearing of AB. i.e. /8 = - a or a = 6 - /8 , i.e. Old bearing - New bearing (3.28) The new co-ordinates of B may now be computed: E B = Ejj + s sin/S = E A + s sin 6 cos a - s cos 6 sin a Efl = ^A + &&AB cos a - &^AB sin a (3.29) Similarly, Nfl = NJ^ + s cos/8 CO-ORDINATES 159 then, and = N A + s cos 6 cos a + s sin 6 sin a N'b = N^ + AN^ cos a + AE^sina (3.30) If a scale factor k is required (e.g. to convert feet into metres), E* = E 'a + AE^ = E^ + ZctAE^gCos a - AN^ sin a] K = Ni + ANJS i 5 = l^ + iiiMfl = N^ + /ctAN^ cos a + AE^sina] From the above, AE^ B = fctAE^ cos a - AN^ sin a] = mAE^ - nAN^ AN^ S = mAN^ + nAE AB where m = k cos a and n = k sin a If the angle of rotation (a) is very small, the equations plified as cosa-»0 and sin a -> a radians. E5 = V A + k[AE AB -, fiN AB a] N B = N^ + tfAN^ + AE^ a] Example 3.16 Transposition of grid (3.31) (3.32) (3.33) (3.34) are sim- (3.35) (3.36) Fig. 3. 27 160 SURVEYING PROBLEMS AND SOLUTIONS In Fig. 3.27, AE AN Let OA = 045° i.e. N45°E 400 + 282-84 + 282-84 OB = 120° S60°E 350 + 303-10 - 175-00 OC = 210° S30°W 350 -175-00 -303-10 OD = 330° N30°W 400 -200-00 + 346-40 If the axes are now rotated through -15 c ' the bearings will be in creased by +15°. AE' AN' OA' = 060° N60°E 400 + 346-40 + 200-00 OB' = 135° S45°E 350 + 247-49 -247-49 OC' = 225° S45°W 350 + 247-49 -247-49 OD' = 345° N15°W 400 -103-52 + 386-36 Applying the transposition of the grid formulae; equation (3.29) AE' = AEcosa- AN sina equation(3.30) AN' = AN cos a + AE sin a AE An Ae cosa AN sina An cosa Ae sina Ae' OA OB OC OD + 282*84 + 303.10 - 175-00 -200-00 + 282-84 -175-00 -303.10 + 346-40 +273-20 + 292-77 -169-04 -193-19 -73-20 + 45-29 + 78-45 -89-68 + 273-20 - 169-04 -292-77 + 334-60 -73-20 -78-45 +45-29 + 51-76 + 346-40 + 247-48 -247-49 -103-51 An' + 200-00 -247-49 -247-48 + 386-36 N.B. (1) cos (-15°) - +0-965 93 (2) sin (-15°) = -0-258 82 (3) If the point of rotation (slew) had a co-ordinate value (EqNq) based on the new axes, these values would be added to the par- tial values, AE'AN' to give the new co-ordinate values. 3.7 The National Grid Reference System Based on the Davidson Committee's recommendations, all British Ordnance Survey Maps will, on complete revision, be based on the National Grid Reference System with the metre as the unit. The origin of the 'Modified Transverse Mercator Projection' for the British Isles is Latitude 49° N Longitude 2° W To provide positive co-ordinates for the reference system a 'False Origin' was produced by moving the origin 100 km North and 400 km West . The basic grid is founded upon a 100 km square; commencing from the false origin which lies to the S.W. of the British Isles, and all squares are referenced by relation to this corner of the square. CO-ORDINATES 161 /0 100 200 300 400 500 600 700km E False origin _1_ 49° N latitude True -^T 2* W long'itucle True origin Fig. 3.28 Old O.S. grid reference system 162 SURVEYING PROBLEMS AND SOLUTIONS 'Eastings are always quoted first.' Originally the 100 km squares were given a reference based on the number of 100 kilometres East and North from the origin (see Fig. 3.28). Subsequently, 500 km squares were given prefix letters of S, N and H, and then each square was given a letter of the alphabet (neglecting I). To the right of the large squares the next letter in the alphabet gives the appropriate prefixes, T, O and J (see Fig. 3.29). km ► 1 Central meridian L M N o P L M H- Hebrides Q R S T U Q 1 R J V W X Y z V W A B C D E A B F G H J K F G ■ 800 N- North L M O P L M Q R s S J T U Q L-*- V W X Y z V w A B c D E A B F G H J K F G -ir 300 S- South L M -N- \ O P L T II Q .' U a u R S V W X Y z V . / False origin -?#— 49* N latitude True -^"^ < . origin -^ 2 W longitude Fig. 3.29 New O.S. grid reference system Square 32 becomes SO 43 becomes SK 17 becomes NM The basic reference map is to the scale 1/25000 (i.e. approxi- mately 2y 2 inches to 1 mile), Fig. 3.30. Each map is prefixed by the reference letters followed by two digits representing the reference numbers of the SW corner of the sheet. See example (Fig. 3.30), i.e. SK54. This shows the relation- ship between the various scaled maps and the manner in which each sheet is referenced. CO-ORDINATES 163 A point P in Nottingham Regional College of Technology has the grid co-ordinates E 457 076-32 m, N 340 224-19 m. Its full 'Grid Refer- ence' to the nearest metre is written as SK/5740/076 224 and the sheets on which it will appear are: Reference Scale SK 54 1/25 000 ( ^ 2*/ 2 in. to 1 mile) SK 54 SE 1/10 560 ( 6 in. to 1 mile) SK 57 40 1/2 500 ( ^ 25 in. to 1 mile) SK 57 40 SW 1/1 250 ( 2; 50 in. to 1 mile) Sheet size Grid size 10 km 1km 5 km 1km 1km 100 m 500 m 100 m 10km 350 000 49 48 46 45 340000 SK 5740 SW (1/1250) 5km SK 54 (1/25000) SK 54 SE (6" to imile) SK 5740 (1/2500) $5 -* "» u>A oi O) noo» <o o» o o o 8 ° o Fig. 3.30 O.S. sheet sizes 8 Exercises 3(d) (Co-ordinates) 16. The co-ordinates of stations A and B are as follows: Latitude Departure A +8257 m + 1321m S +7542 m - 146 m Calculate the length and bearing of AB (Ans. 244° 01'; 1632 m) 17. The co-ordinates of two points A and B are given as: A N 188 -6 m E 922-4 m B S 495-4 m E 58-6 m 164 SURVEYING PROBLEMS AND SOLUTIONS Calculate the co-ordinates of a point P midway between A and B. (Ans. S 153-4 m, E 490-5 m) 18. The bearings of a traverse have been referred to the magnetic meridian at the initial station A and the total co-ordinates of B, rela- tive to A, are found to be 368 m W, 796 m S. Calculate (a) the length and magnetic bearing of AB, (b) the true bearing of AB assuming that the magnetic declination is 13° 10' W of true north, (c) the co-ordinates of B with reference to the true meridian at the initial station. (Ans. (a) AB mag. bearing 204° 49' 877 m (b) true bearing 191° 39'; (c) corrected co-ordinates 177-1 W, 858-9 S) 19. The co-ordinates, in metres of two points, X and Y, are as follows: X W 582-47 m N 1279-80 m Y E 1191-85 m S 755-18 m Calculate the length and bearing of XY. (Ans. 2699-92 m; 138°54'50") 20. Survey station X has a Northing 424-4 ft, Easting 213*7 ft and a height above Ordnance datum of 260*8 ft. Station Y has a Northing 1728-6 ft, Easting 9263-4 ft and a depth below Ordnance datum 763*2 ft. Find the length, bearing and inclination of a line joining XY . (Ans. 9143-3 ft; 081° 47' 54"; 1 is 8-92) 21. The co-ordinates of four survey stations are given below: Station North (ft) East (ft) A 718 90 B 822 469 C 164 614 D 210 81 Calculate the co-ordinates of the intersection of the lines AC and BD. (L.U. Ans. N 520, E 277) 22. Readings of lengths and whole circle bearings from a traverse carried out by a chain and theodolite reading to 1 minute of arc were as follows, after adjusting the angles: Line AB BC CD DE W.C.B. 0°00' 35° 40' 46° 15y 2 ' 156° 13' Length (ft) 487-2 538*6 448-9 295-4 Line EF FA AG GH HD W.C.B. 180° 00' 270° 00' 64° 58' 346° 25' 37° 40' Length (ft) 963-9 756-2 459-3 590-7 589-0 CO-ORDINATES 165 Taking the direction AB as north, calculate the latitude and de- parture of each line. If A is taken as origin and the mean co-ordinates of D as obtained by the three routes are taken as correct, find the co- ordinates of the other points by correcting along each line in propor- tion to chainage (answers are required correct to the nearest 0*1 ft) (L.U. Ans. D = 1234-7 ft N, 637-6 ft E) 23. The following notes were taken during a theodolite traverse: Bearing of line AB 14° 48' 00" Angle observed Length (metres) ABC 198° 06' 30" AB 245 BCD 284° 01 '30" BC 310 CDE 200° 12' 30" CD 480 DEF 271° 33' 30" DE 709 EFG 268° 01 '30" EF 430 FG 607 Calculate the length and bearing of the line GA. (Ans. 220-6 m; N 61° 27' 40" W) 24. From the following notes, calculate the length and bearing of the line DA: Line Bearing Length AB 015° 30' 630 m BC 103° 45' 540 m CD 270° 00' 227 m (Ans. 668 m; S 44° 13' W) 25. The notes of an underground traverse in a level seam are as follows: Line Azimuth Distance (ft) AB 30° 42' BC 86° 24' 150-6 CD 32° 30' 168-3 DE 315° 06' 45-0 The roadway DE is to be continued on its present bearing to a point F such that F is on the same line as AB produced. Calculate the lengths of EF and FB. (M.Q.B./M Ans. EF 88-9 ft; FB 286-2 ft) 26. A shaft is sunk to a certain seam in which the workings to the dip have reached a level DE. It is proposed to deepen the shaft and connect the point E in the dip workings to a point X by a cross- measures drift, dipping at 1 in 200 towards X. The point X is to be Line Azimuth AB 270° 00' BC 184° 30' CD 159° 15' DE 90° 00' 166 SURVEYING PROBLEMS AND SOLUTIONS 134 ft from the centre of the shaft A and due East from it, AX being level. The following ate the notes of a traverse made in the seam from the centre of the shaft A to the point E. Distance Vertical Angle 127 Level 550 Dipping 21° 730 Dipping 18y 2 ° 83 Level Calculate (a) the azimuth and horizontal length of the drift EX and (b) the amount by which it is necessary to deepen the shaft. (M.Q.B./M Ans. (a) 358° 40' 1159-6 ft (b) 434-5 ft) 27. The notes of a traverse between two points A and £ in a certain seam are as follows: Distance (ft) Angle of Inclination 600 +6° 450 -30° 550 level 355 level It is proposed to drive a cross-measures drift from a point E to another point F exactly midway between A and B. Calculate the azimuth and length EF. (M.Q.B./M Ans. 359° 33'; 867 ft, 888-3 ft inclined) 28. Undernoted are details of a short traverse between the faces of two advancing headings, BA and DE, which are to be driven forward until they meet: Line Azimuth Distance AB 80° 270-6 ft BC 180° 488-0 ft CD 240° 377-0 ft DE 350° 318-0 ft Calculate the distance still to be driven in each heading. (M.Q.B./M Ans. BA + 168-4 ft; DE + 291-5 ft) 29. In order to set out the curve connecting two straights of a road to be constructed, the co-ordinates on the National Grid of /, the point of intersection of the centre lines of the straights produced, are re- quired. L,ine Azimuth AB 89° BC 170° CD 181° DE 280° CO-ORDINATES 167 A is a point on the centre line of one straight, the bearing Al being 72° 00' 00", and B is a point on the centre line of the other straight, the bearing IB being 49° 26' 00" . Using the following data, calculate with full checks the co-ordinates of/. Eastings (ft) Northings (ft) A +43758-32 +52202-50 B +45165-97 +52874-50 The length AB is 1559-83 ft and the bearing 64° 28' 50". (N.U. Ans. E + 45 309-72 N + 52706-58) 30. It is proposed to sink a vertical shaft to connect X on a roadway CD in the upper horizon with a roadway GH in the lower horizon which passes under CD. From surveys in the two horizons the following data are compiled: Upper horizon Station Horizontal Angle Inclination Inclined Length Remarks A co-ordinates of A + 1 in 200 854-37 E 6549-10 ft B 276° 15' 45" N 1356-24 ft + 1 in 400 943-21 Bearing AB C 88° 19' 10" N 30° 14' 00" E. Level D Lower horizon Station Horizontal Angle Inclination 736-21 Inclined Length + lin50 326-17 G H 193° 46' 45" 83° 03' 10" + 1 in 20 level 278-66 626-10 Remarks Co-ordinates of £ E 7704-08 ft N 1210-88 ft Bearing EF N 54° 59' 10" E. Calculate the co-ordinates of X (Ans. E 8005*54 ft, N 1918*79 ft) 31 . The surface levels of two shafts X and Y and their depth are respectively as follows: Surface Level Depth X 820-5 ft A.O.D. 200yd Y 535*5 ft A.O.D. 150 yd 168 SURVEYING PROBLEMS AND SOLUTIONS The co-ordinates of the centre of the two shafts in fact, are re- spectively as follows: E N X -778-45 +2195-43 Y +821*55 + 359-13 Calculate the length and gradient of a cross-measures drift to con- nect the bottom of the shaft. (Ans. 2439-3 ft(incl.), 2435-6 ft(hor.); 3° 10', i.e. 1 in 18) 32. The co-ordinates of A are N25m E13m. From ,4 a line AB runs S44°ll' E for 117m. On the line AB an equilateral triangle ABC is set out with C to the north of AB. Calculate the co-ordinates of B and C. (Ans. B E + 94-5m, N-55-9m. C E+ 126-4 m, N + 53-7m) 33. (a) Calculate the gradient (as a percentage) between two points, M and N, which have been co-ordinated and heighted as given below: p • t Co-ordinates Height E(ft) N(ft) (ft) M 6206-5 3465-2 212-4 # 5103-2 2146-8 196-6 6002-5 2961-4 (b) Determine the length (in centimetres) of the line MN when plotted at a scale 1 : 500 (assume 1 f t = 0-3048 m). (c) Calculate the bearing of the line MO (R.I.C.S. Ans. 0-92%; 104-8cm;202°03') 34. From an underground traverse between 2 shaft wires A and D the following partial co-ordinates in feet were obtained: AB E 150-632 ft, S 327 -958 ft BC E 528-314 ft, N 82-115 ft CD E 26-075 ft, N 428-862 ft Transform the above partials to give the total Grid co-ordinates of station B given that the Grid co-ordinates of A and D were: A E 520 163-462 metres, N 432 182-684 metres D E 520378-827 metres, N 432238-359 metres (Aide memoire) X = x, + K(x -yd) Y = y, + K(y + xd) (N.R.C.T.) Bibliography MINISTRY OF DEFENCE, Textbook of Topographical Surveying, 4th ed. (H.M.S.O.) COORDINATES 169 GLENDENNING, J., Principles of Surveying (Blackie) CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) HOLLAND, J.L., WARDELL, K. and WEBSTER, A. C, Surveying, Coal- mining Series, (Virtue) A Brief Description of the National Grid and Reference System. O.S. Booklet No. 1/45 (H.M.S.O.) middleton and CHADWICK, Treatise on Surveying, Vol. 1 (Spon). salmon, V.G., Practical Surveying and Field Work (Griffin) BRINKER, R.C. and TAYLOR, W.C., Elementary Surveying, 4th ed. (International Textbook Co.) Appendix Comparison of scales Scales in common use with the metric system Scales in common use with the foot/inch system Recommended Other Alternative byBSI Scales i : ioooooo \ 1 ioooooo A in to 1 mile approx. I : 500000 1 : 625000 1 : 625000 Id in to 1 mile approx. 1 : 200000 1 : 250000 1 : 250000 i in to 1 mile approx. 1 : 100000 1 : 125000 1 : 126720 \ in to 1 mile 1 : 50000 1 : 62500 1 : 63360 1 in to 1 mile 1 : 20000 1 : 25000 1 : 25000 2 J in to 1 mile approx. 1 : 10000 1 : 10560 6 in to 1 mile 1 : 5000 1 : 2000 1 : 2500 1 : 2500 1 in to 208.33 ft 1 : 1000 1 : 1250 1 : 1250 1 in to 104.17 ft 1 : 500 1 : 500 1 in to 41.6 ft 1 :384 A in to r ft 1 : 200 1 : 192 ■is in to 1 ft 1 : 100 1 : 9 6 J in to 1 ft 1 :so 1 : 4 8 i in to 1 ft 1 : 20 1 :24 \ in to 1 ft 1 : 10 1 : 12 1 in to 1 ft 1 :5 1 = 4 3 in to 1 ft From Chartered Surveyor, March, 1968. 4 INSTRUMENTAL OPTICS 4.1 Reflection at Plane Surfaces 4.11 Laws of reflection (1) The incident ray, the reflected ray, and the normal to the mirror at the point of incidence all lie in the same plane. (2) The angle of incidence (i) = the angle of reflection (r). A ... 8 C Fig. 4.1 The ray AO, Fig. 4. 1, is inclined at a (glancing angle MO A) to the minor MN. Since i = r, angle BON = MO A = a. If AO is pro- duced to C, Angle MO A = NOC = BON = a Thus the deviation of the ray AO is 2a. Therefore the deviation angle is twice the glancing angle, i.e. D = 2a (4.1) 4.12 Deviation by successive reflections on two inclined mirrors (Fig. 4.2) Ray AB is incident on mirror A^/V, at a glancing angle a. It is thus deflected by reflection + 2a . The reflected ray BC incident upon mirror M 2 N Z at a glancing angle j8 is deflected by reflection - 2j8 (here clockwise is assumed +ve). 170 INSTRUMENTAL OPTICS 171 \ Fig. 4.2 The total deflection D is thus (2a- 2/3) = 2(a-j8). In triangle BCX, )8 = a + =0-a i.e. D = 2(a-j8) = 20 (4.2) As is constant, f/ie deflection after two successive reflections is constant and equal to twice the angle between the mirrors. 4.13 The optical square (Fig. 4.3) This instrument, used for setting out right angles, employs the above principle. By Eq. (4.2), the deviation of any ray from 2 incident on mirror M 2 at an angle a to the normal = 20, i.e. 2 x 45° = 90°. 4.14 Deviation by rotating the mirror (Fig. 4.4) Let the incident ray AO be constant, with a glancing angle a. The mirror M, N, is then rotated by an anticlockwise angle /3 to MgN^. When the glancing angle is a the deviation angle is 2a. After rotation the glancing angle is (a+ /3) and the deviation angle is therefore 2(a+ /S ) . Thus the reflected ray is rotated by <f> = 2(a+£)-2a = 2/8 (4.3) 172 SURVEYING PROBLEMS AND SOLUTIONS // the incident ray remains constant the reflected ray deviates by twice the angular rotation of the mirror. Fig. 4.3 Optical square Fig. 4.4 4.15 The sextant Principles of the sextant (Figs. 4.5 and 4.6) Mirror M, , silvered, is connected to a pointer P. As M, is rotated the pointer moves along the graduated arc. Mirror M 2 is only half-silvered and is fixed. When the reading at P is zero, Fig. 4.5, the image K, reflected from both mirrors, should be seen simultaneously with K through the plain glass part of M 2 . With a suitable object K as the horizon, the mirrors should be parallel. INSTRUMENTAL OPTICS 173 From triangle M y EM z , from triangle M^QM 2 , Fig. 4.5 Zero setting on the sextant When observing an elevated object S, Fig. 4.6, above the horizon K, the mirror M, is rotated through angle <j> until S is simultaneously observed with K. The angle being measured is therefore 6. 2a = 2/3 + 6 6 = 2(a-0) 90 + a = 90 + p + <f> a = j8 + <j6 4 = a-j8 = £0 (4.4) i.e. f/ie rotation of the mirror is half the angle of elevation. When the mirrors are parallel, <f> = with the index pointer MP at zero on the graduated arm. When the angle 6 is being observed, the mirror is turned through angle 0, but the recorded value MP 2 = 6, i.e. 2<f>. N.B. (1) Horizontal angles are only measured if the objects are at the same height relative to the observer, which means that in most cases the angle measured is in an inclined plane. The horizontal angle may be computed from the equation (see page 107): cos - cos a, cos a. cos H.A. = i sin a, sina 2 where 6 = the measured angle in the inclined plane and a, and a 2 = vertical angles. (2) Vertical angles must be measured relative to a true or an artificial horizon. 174 SURVEYING PROBLEMS AND SOLUTIONS s Recorded value 8 Fig. 4.6 Principles of the sextant 4.16 Use of the true horizon (a) As the angle of deviation, after two successive reflections, is independent of the angle of incidence on the first mirror, the object will continue to be seen on the horizon no matter how much the observer moves. Once the mirror W, has been set, the angle between the mirrors is set, and the observed angle recorded. This is the main advantage of the sextant as a hand instrument, particularly in marine and aerial navigation where the observer's posi- tion is unstable. (b) If the observer is well above the horizon, a correction 50 is required for the dip of the horizon, Fig. 4.8. INSTRUMENTAL OPTICS 175 \ Fig. 4. 7 Box sextant The Nautical Almanac contains tables for the correction factor 86 due to the dip of the horizon based on the equation: 86 = -0-97 yjh minutes where h = height in feet above sea level, or 86 = -1-756 V^ minutes where H = height in metres above sea level. (4.5) Fig. 4.8 Dip of the horizon 4.17 Artificial horizon (Fig. 4.9) On land, no true horizon is possible, so an 'artificial horizon' is employed. This consists essentially of a trough of mercury, the surface of which assumes a horizontal plane forming a mirror. 176 SURVEYING PROBLEMS AND SOLUTIONS £-|> ^^J Mercury plane Fig. 4.9 Artificial horizon The vertical angle observed between the object S and the reflec- tion of the image S, in the mercury is twice the angle of altitude (a) required. Observed angle = S,ES = 2a True altitude = MS, S = a Rays SE and SS^ are assumed parallel due to the distance of S from the instrument. 4.18 Images in plane mirrors Image Fig. 4. 10 Images in plane mirrors Object in front of the mirror is seen at E as though it were situated at /, Fig. 4.10. From the glancing angles a and /S it can be seen that (a) triangles OFC and ICF are congruent, (b) triangles OFD and IDF are congruent. Thus the point / (image) is the same perpendicular distance from the mirror as (object), i.e. OF = FI. INSTRUMENTAL OPTICS 177 4.19 Virtual and real images As above, the rays reflected from the mirror appear to pass through /, the image thus being unreal or virtual. For the image to be real, the object would have to be virtual. The real test is whether the image can be received on a screen: if it can be— it is real, if not— it is virtual. 4.2 Refraction at Plane Surfaces Normal /= Angle of incidence r» Angle of refraction Boundary of media Refracted ray Fig. 4.11 The incident ray AO, meeting the boundary between two media, e.g. air and glass, is refracted to B, Fig. 4.11. 4.21 Laws of refraction (1) The incident ray, the refracted ray, and the normal to the bound- ary plane between the two media at the point of incidence all lie in the same plane. (2) For any two given media the ratio — . is a constant known as sin r the refractive index (the light assumed to be monochromatic). sin i Thus Refractive Index = sin r (4.6) 4.22 Total internal reflection (Fig. 4.12) If a ray AO is incident on a glass/ air boundary the ray may be refracted or reflected according to the angle of incidence. 178 SURVEYING PROBLEMS AND SOLUTIONS (b) Critical angle (C) Reflected Fig. 4.12 When the angle of refraction is 90°, the critical angle of incidence is reached, i.e. sin c . /a n\ rMa = TTTono = sin c (4.7) g' = sin c 1-5 For crown glass the refractive index a fi g •' sinc-^g c ^ 41° 30' If the angle of incidence (glass/ air) i : > 41° 30' , the ray will be internally reflected, and this principle is employed in optical prisms within such surveying instruments as optical squares, reflecting prisms in binoculars, telescopes and optical scale-reading theodolites. N.B. Total internal reflection can only occur when light travels from one medium to an optically less dense medium, e.g. glass/ air. 4.23 Relationships between refractive indices (Fig. 4.13) (a) // the refractive index from air to glass is a fj, g , then the re- fractive index from glass to air is g fM a Therefore e fjL a = e.g., if a /J. g = 1-5 (taking air as 1), then (4.8) 8 u„ = — = 0-66 ^ a 1-5 _ (b) Given parallel boundaries of air, glass, air, then sin i = constant sin i a (4.9) /** = -Ma = sin i 8 sin i sin i, aPi gfia sin i a = a fi g sin i g = constant INSTRUMENTAL OPTICS 179 Fig. 4.13 (c) The emergent ray is parallel to the incident ray when returning to the same medium although there is relative displacement. This factor is used in the parallel plate micrometer. 4.24 Refraction through triangular prisms When the two refractive surfaces are not parallel the ray may be bent twice in the same direction, thus deviating from its former direc- tion by an angle D. It can be seen from Fig. 4.14 that A = ft + ft (4.10) and D = (a, - ft) + (cl,- ft) (4.11) = (a 1+ a 2 )-(ft + ft) (4.12) i.e. D = (a, + a 2 ) - A (4.13) Thus the minimum deviation occurs when a, + Og = A (4.14) If A is small, then a = /x/3 and D = /ift + /xft -A = Mft+ft)-^ = AQi-1) (4.15) 180 SURVEYING PROBLEMS AND SOLUTIONS Fig. 4.14 Refraction through a triangular prism 4.25 Instruments using refraction through prisms The line ranger (Fig. 4.15) 0^ Prism P 2 Prism P^ Field of view Fig. 4.15 The line ranger a+)8 = 90° .'. 2(a+ /3) = 180° Thus 0, C0 2 is a straight line. INSTRUMENTAL OPTICS 181 The prism square (Fig. 4.16) ^-~* Fig. 4.16 The prism square This is precisely the same mathematically as the optical square (Fig. 4.3), but light is internally reflected, the incident ray being greater than the critical angle of the glass. The double prismatic square (Fig. 4.17) combines the advantages of both the above hand instruments. Fig. 4.17 182 SURVEYING PROBLEMS AND SOLUTIONS Images 2 and 3 are reflected through the prisms. 0, is seen above and below the prisms. The parallel plate micrometer (Fig. 4.18) t Fig. 4.18 The parallel plate micrometer A parallel- sided disc of glass of refractive index /a and thickness t is rotated through an angle 6. Light is refracted to produce displace- ment of the line of sight by an amount x. x = DB = AB sin(0 - <f>) = — — : x sin(0 - d>) cos f(sinfl cos<ft - cosfl sin<ft) ~ cos<£ = f(sin# - cos0 tan</>) but refractive index fj. = sin0 = sin# sin 4> sinfl and cos<£ = y/\l - sin 2 <£} vV 2 -sin 2 0} -<[ INSTRUMENTAL OPTICS cos0 sin# 1 183 sin0 - = t sin0 = t sin 6 1- VOtx 2 - sin 2 0). cos0 VO? -sin 2 0) If is small, then sin0~0 rad and sin 2 may be neglected. .-. x * ifl(l-i) (4.17) Example 4.1 A parallel plate micrometer attached to a level is to show a displacement of 0*0 1 when rotated through 15° on either side of the vertical. Calculate the thickness of glass required if its refractive index is 1-6. State also the staff reading to the nearest thousandth of a foot when the micrometer is brought to division 7 in sighting the next lower read- ing of 4*24, the divisions running to 20 with 10 for the normal posi- tion. (L.U.) Using the formula x = t sin 'Hfc£h)\ t = siae h . // (l-s^)(l + si,^ 1 [ *J \(jjl - sin 0)Qi + sin &))} 0-01 x 12 in. sin icoN _ If (l-sinl5)(l + sinl5) )] L a/ 1(1-6 -sin 15) (1-6+ sin 15)/ J 0-01 x 12 0-25882 x 0-388 24 in ' = 1-1940 in. N.B. If the approximation formula is used, t = 1-222 in. The micrometer is geared to the parallel plate and must be corre- lated. Precise levelling staves are usually graduated in feet and fiftieths of a foot, so the micrometer is also divided into 20 parts, each representing 0*001 ft. (The metric staff requires a metric micro- meter). To avoid confusion, the micrometer should be set to zero before each sight is taken and the micrometer reading is then added to the staff reading as the parallel plate refracts the line of sight to the next lower reading. 184 SURVEYING PROBLEMS AND SOLUTIONS Z* a 3 ZP !R - • -» - s -* \ Reading 4-24 Staff 0-0070 Micrometer 4-2470 Fig. 4.19 Use of the parallel plate micrometer in precise levelling Exercises 4(a) 1. Describe the parallel plate micrometer and show how it is used in precise work when attached to a level. If an attachment of this type is to give a difference of 0-01 of a foot for a rotation of 20°, calculate the required thickness of glass when the refractive index is 1*6. Describe how the instrument may be graduated to read to 0-001 of a foot for displacements of 0-01 of a foot above and below the mean. (L.U. Ans. 0-88 in.) 2. Describe the method of operation of a parallel plate micrometer in precise levelling. If the index of refraction from air to glass is 1*6 and the parallel plate prism is 0*6 in. thick, calculate the angular rotation of the prism to give a vertical displacement of the image of 0*001 ft. (L.U. Ans. 3° 03' 36") 4.3 Spherical Mirrors 4.31 Concave or converging mirrors (Fig. 4.20) A narrow beam of light produces a real principal focus F. P is called the pole of the mirror and C is the centre of curvature. PF is the focal length of the mirror. A ray AB, parallel to the axis, will be reflected to F. BC will be normal to the curve at B, so that Angle ABC = As AB is parallel to PC, Angle PCB = BF = angle CBF = 6. angle ABC = 6 FC. INSTRUMENTAL OPTICS 185 Fig. 4.20 Concave mirror As the beam is assumed narrow PF ^ BF ~ FC :. PC ^ 2PF = 2/ ^ r. (4.18) If the beam of light is wide a cusp surface is produced with the apex at the principal focus. The parabolic mirror overcomes this ano- maly and is used as a reflector for car headlights, fires, etc, with the light or heat source at the focus. Fig. 4.21 Wide beam on a circular mirror 186 SURVEYING PROBLEMS AND SOLUTIONS Fig. 4.22 Parabolic mirror 4.32 Convex or diverging mirrors (Fig. 4.23) C^" Fig. 4.23 Convex mirror A narrow beam of light produces a virtual principal point F, being reflected away from the axis. The angular principles are the same as for a concave mirror and r e 2/ 4.33 The relationship between object and image in curved mirrors Assuming a narrow beam, the following rays are considered in all cases (Fig. 4.24). (a) Ray OA, parallel to the principal axis, is reflected to pass through the focus F. (b) Ray OB, passing through the focus F, is then reflected parallel to the axis. (c) Ray OD, passing through the centre of curvature C, and thus a line normal to the curve. N.B. In graphical solutions, it is advantageous to exaggerate the vertical scale, the position of the image remaining in the true position. As the amount of curvature is distorted, it should be represented as a INSTRUMENTAL OPTICS 187 straight line perpendicular to the axis. Any two of the above rays produce, at their intersection, the posi- tion of the image /. \ ^ \^i* / '! ^^r\. /C Oy 1 /*\ l ! % Fig. 4.24 The relationships between object and image for concave mirrors are: (a) When the object is at infinity, the image is small, real, and inverted. (b) When the object is at the centre of curvature C, the image is also at C, real, of the same size and inverted. (c) When the object is between C and F, the image is real, en- larged and inverted. (d) When the object is at F, the image is at infinity. (e) When the object is between F and P, the image is virtual, en- larged and erect. For convex mirrors, in all cases the image is virtual, diminished and erect, Fig. 4.25. Fig. 4.25 4.34 Sign convention There are several sign conventions but here the convention Real- is-positive is adopted. This has many advantages provided the work is not too advanced. 188 SURVEYING PROBLEMS AND SOLUTIONS All real distances are treated as positive values whilst virtual distances are treated as negative values— in all cases distances are measured from the pole. N.B. In the diagrams real distances are shown as solid lines whilst virtual distances are dotted. 4.35 Derivation of Formulae Concave mirror (image real), Fig. 4.26 Object „ c r v b f r 1 |, u 1 1 2 / r " / Fig. 4.26 1 1 = -+ - U V To prove: where / = the focal length of the mirror r = the radius of curvature u = the distance of the object from the pole P v = the distance of the image from the pole P The ray OA is reflected at A to AI making an equal angle a on either side of the normal AC. From Fig. 4.26, 6 = a+ fi .'. a = 6-13 and (f> = 2a + j8 = 209-j3) + i 8 = 2B-P i.e. <£+j8= 26. As the angles a, /3, 6 and <f> are ^ small, B is closely adjacent to to P. <f>rad ~ Sin< £ = Jp P rad ■£! Sin j8 = OP 6 . - sin0 = rad h CP h h 2h 1P + 0P ~ CP I is real so IP is +ve. is real so OP is +ve. i.e. 1 + 1 = 2 v u r INSTRUMENTAL OPTICS ■ J < as f=T ) 189 (4.19) Concave mirror (image virtual), Fig. 4.27 Fig. 4.27 From Fig. 4.27, = £ - a ••• a = j6 - and <f> = 2a - j8 = 208-0-0 = -20 + /8 /. 20 = j8-0 As before, 2fc _ _ft _^_ CP~ OP /P i.e. u v but the image is virtual, therefore v is negative. 2 11 1 r ~ f u v Convex mirror, Fig. 4.28 Fig. 4.28 <f> = a + ••• a = <f> - 6 2a = + /S .-. 2(<£-0) = </> + j3 i.e. <f> = 2(9 + /S <£-j8 = 20 (4.19) 190 SURVEYING PROBLEMS AND SOLUTIONS As before, Thus trad ~ Siti< f> h ~ -IP (/ is virtual .*. IP is -ve.) Prad ~ sin/3 h ~ OP (0 is real OP is +ve). . ^ sin0 rad h ~ -PC (C is virtual ••• PC is -ve). h _ h _ 2h IP OP -PC 1 1 i.e. ^- - — -v u 2_ -r JL = _L r f Therefore, using the sign convention, the formula is common to both types of mirror in all cases. 1 1 — + — v u (4.19) 4.36 Magnification in spherical mirrors (Fig. 4.29) u Fig. 4.29 Magnification in spherical mirrors IB is the image of OA. In the right-angled triangles OPO^ and /P/, the angle a is common, being the angles of incidence and of reflection, and therefore the tri- angles are similar. ... A . /7 i (image size) /, P (v) Thus, magnification = 00^ (object size) 0, P (w) m = — neglecting signs (4.20) INSTRUMENTAL OPTICS 191 Example 4.2 An object 1 in. high is placed on the principal axis 20 in. from a concave mirror which has a radius of curvature of 15 in. Find the position, size and nature of the image. As the mirror is concave, / = the object is real Substituting in Eq. (4.19), 15. + 20 i-I+I J U V 1 = 1-1 V / u 2 1 1 "15 20" 12 •'. v = 12 in. Thus the image is real (but will be inverted) as v is positive. Magnification v m = — u Size of image = 0*6 in. -S-- 4.4 Refraction Through Thin Lenses 4.41 Definitions (a) Types of lens Convex (converging), Fig. 4.30 (a) Concave (diverging), Fig. 4.30(b) Double convex Piano- convex (a) Convex meniscus Double concave Plano-concave Concave meniscus (b) Fig. 4.30 Types of lens 192 SURVEYING PROBLEMS AND SOLUTIONS (b) Focal points (Fig. 4.31) Convex lens (pole) "1 Princ i pal axi s Concave lens Fig. 4.31 Conjugate foci 4.42 Formation of images (Fig. 4.32) If a thin lens is assumed to be split into a series of small prisms, any ray incident on the face will be refracted and will deviate by an angle D = A(/x -1) (Eq.4.15) Fig. 4.32 Formation of images N.B. The deviation angle D is also related to the height h and the focal length /, i.e. D = h/f (4.21) INSTRUMENTAL OPTICS 193 4.43 The relationship between object and image in a thin lens The position of the image can be drawn using three rays, Fig. 4.33. Fig. 4. 33 N.B. Two principal foci, F, and F 2 , exist. (a) Ray OA parallel to the principal axis is refracted to pass through principal focus F 2 . (b) Ray OB passes through the principal focus F, and is then refracted parallel to the principal axis. (c) Ray OPI passes from object to image through the pole P without refraction. Convex lens (a) When the object is at infinity, the image is at the principal focus F 2 , real and inverted. (b) When the object is between infinity and F, , the image is real and inverted. (c) When the object is between F, and P, the image is virtual, magnified, and erect, i.e. a simple magnifying glass. Concave lens The image is always virtual, erect and diminished. 4.44 Derivation of formulae The real-is-positive sign convention is again adopted, but for con- vex lenses the real distances and focal lengths are considered posi- tive, whilst for concave lenses the virtual distances and focal lengths are considered negative. As with mirrors, thin lens formulae depend on small angle approxi- mations. Convex lens (a) Image real, Fig. 4.34 D = a + (3 194 SURVEYING PROBLEMS AND SOLUTIONS By Eq. (4.21), D = 1 1 = 1+ - U V .... i =± + L U V (b) Image virtual, i.e. object between F and P, Fig. 4.35. Fig. 4. 35 D = a - /3 h. = A_A / U V but v is virtual, i .e. negative. i.e. Concave lens (Fig. 4.36) 4 -i + i / u v D = ft - a i.e. INSTRUMENTAL OPTICS f V u but v and / are negative, being virtual distances / u v 195 Fig. 4. 36 4.45 Magnification in thin lenses (Fig. 4.37) Fig. 4. 37 As with spherical mirrors, OPO^ and /P/, are similar right-angled triangles with angle a commpn. //i (image size) OOi (object size) UP (image distance v) magnification m = _L y P (object distance u) — as before 196 SURVEYING PROBLEMS AND SOLUTIONS N.B. This should not be confused with angular magnification or magni- fying power (M), which is defined as the angle subtended at the eye by the image the angle subtended at the eye by the object For the astronomical telescope, with the image at infinity, M = focal length of objective _ jo focal length of eyepiece f e (4.22) 4.5 Telescopes 4.51 Kepler's astronomical telescope (Fig. 4.38) Objective Eyepiece Diaphragm Fig. 4.38 Kepler's atronomical telescope The telescope is designed to increase the angle subtending dis- tant objects and thus apparently to bring them nearer. The objective lens, converging and of long focal length, produces an image FX, inverted but real, of the object at infinity. The eyepiece lens, converging but of short focal length, is placed close to F so as to produce from the real object FX a virtual image IY , magnified but similarly inverted. 4.52 Galileo's telescope (Fig. 4.39) The eyepiece is concave and produces a virtual, magnified, but erect image IY of the original inverted image XF produced by the objective. As the latter image lies outside the telescope eyepiece, it is unsuitable for surveying purposes where cross hairs are required. INSTRUMENTAL OPTICS 197 Objective Fig. 4.39 Galileo's telescope 4.53 Eyepieces Ideally, the eyepieces should reduce chromatic and spherical aberration. Lenses of the same material are achromatic if their distance apart is equal to the average of their focal lengths, i.e. d = i(/,+/ 2 ) (4-23) If their distance apart is equal to the differences between their focal lengths, spherical aberration is reduced, i.e. d = /, - U (4.24) For surveying purposes the diaphragm must be between the eye- piece and the objective. The most suitable is Ramsden* s eyepiece, Fig. 4.40. Diaphragm Fig. 4.40 Ramsden's eyepiece The focal length of each lens is the same, namely /. Neither of the conditions (4.23) or (4.24) is satisfied. Chromatic |-(/i+/ 2 ) = / compared with 2/3/ Spherical /, - / 2 = compared with 2/3 / Huyghen's eyepiece, Fig. 4.41, satisfies the conditions but the focal plane lies between the lenses. It is used in the Galileo telescope. 198 SURVEYING PROBLEMS AND SOLUTIONS Diaphragm Fig. 4.41 Huyghen's eyepiece Chromatic condition -j (3/ + f) = 2/ = d Spherical condition 3/ - / = 2/ = d Example 4.3 An astronomical telescope consists of two thin lenses 24 in. apart. If the magnifying power is xl2, what are the focal lengths of the two lenses ? Fig. 4.42 magnifying power = = 12 12/. = h But f + f e = 26 in. /. 12/ e + f e = 26 in. 26 f e = =-r - 2 in. eyepiece lens fo = 12/ e 24 in. objective lens 4.54 The internal focussing telescope (Fig. 4.43) The eyepiece and objective are fixed and an internal concave lens is used for focussing. For the convex lens, by Eq. (4.19) 1 /, 1 + 1 INSTRUMENTAL OPTICS 199 Objective Internal focussing lens Diaphragm i.e. For the concave lens, Fig. 4.43 Internal focussing telescope 1 1 - 1 or 1 u, u, 1 1 «2 = -(f,-<0 1 /a .1+1 "2 V 2 1 u 1 = + v, - d 1 l-d 1 /a 1 v, -d / 1 -d (4.25) An internal focussing telescope has a length / from the objective to the diaphragm. The respective focal lengths of the objective and the internal focussing lens are /, and / 2 . To find the distance d of the focussing lens from the objective when the object focussed is u, from the objective, Fig. 4.43. For the objective, \ 11 For the focussing lens, /a -JU i.e. 1 _ 1 u. i.e. v, - d / - d (v, - d)(/ - d) = / a (/ - d) - / 2 (v, - d) d 2 - dtf + v,) + { / Vl -/ 2 (/-v,)} = d 2 - dC/ + v,) + W/ + / a ) -f 2 l\ = (4.26) 200 SURVEYING PROBLEMS AND SOLUTIONS This is a quadratic equation in d and its value will vary according to the distance «, of the object from the instrument. Example 4.4 Describe, with the aid of a sketch, the function of an internal focussing lens in a surveyors' telescope and state the advant- ages and disadvantages of internal focussing as compared with external focussing. In a telescope, the object glass of focal length 7 in. is located 9 in. away from the diaphragm. The focussing lens is midway between these when the staff 60 ft away is focussed. Determine the focal length of the focussing lens. (L.U.) Internal focussing lens Diaphragm Fig. 4.44 For the convex objective lens, /, = 7 in. m, = 60 x 12 = Then, by Eq. (4.19), 1 - JL - i = 1- JL 7 720 720 - 7 720 in. 720 x For the focussing lens, v, - 4-5 = 7-068 4-5 1 u 2 1 u 713 5040 4-5 = 2-568 + u. u - 2-568 4-5 -4-5 + 2-568 11-556 -5-98 in. (i.e. the lens is concave) INSTRUMENTAL OPTICS 201 Example 4.5 In an internally focussing telescope, Fig. 4.43, the objective of focal length 5 in. is 7*5 in. from the diaphragm. If the in- ternal focussing lens is of focal length 10 in., find its distance from the diaphragm when focussed to infinity. For the objective, /, = 5 in. and thus the position of F, will be 5 in. from C,. C 2 F, = 5 - d For the internal focussing lens, /a = -10 u 2 = -(5-d) v, = 7-5 - d i.e. i.e. 1 = ± + J- f 2 U 2 *2 1 - l . X 10 5 _ d ' 7-5 - d -(5 - <0(7-5 - d) = -10(7-5 - d) + 10(5 - d) (37-5 - 12-5d + d z ) = -75 + lOd + 50 - lOd = -25 d 2 - 12-5d + 12-5 = d = 4-235 in. v 9 = 7-5 - 4-235 = 3-265 in. i.e. the internal focussing lens will be 3-265 in. away from the dia- phragm when focussed to infinity. 4.55. The tacheometric telescope (external focussing) (Fig. 4.45) Staff \fertical axis Diaphragm Fig. 4.45 The tacheometric telescope (external focussing) Let a, b and c represent the three horizontal cross hairs of the diaphragm, ac being a distance i apart and b midway between a and 202 SURVEYING PROBLEMS AND SOLUTIONS With the telescope in focus, these lines will coincide with the image of the staff observed at A, B and C respectively; the distance AC = s is known as the staff intercept. The line bOB represents the line of collimation of the telescope, with bO and OB conjugate focal lengths of the lens, v and u, respectively. The principal focal length of the lens is FO (/), whilst the vertical axis is a distance k from the principal focus F. Because the triangles acO and ACO are similar, AC = QB or ± = u_ ac ob i v Using the lens formula, Eq.(4.19), f u v and multiplying both sides by uf gives, V Substituting the value of u/v from Eq.(4.27), u = s- + / i Thus the distance from the vertical axis to the staff is given as D = sl+ (/+d) (4.28) This is the formula which is applied for normal stadia observations with the telescope horizontal and the staff vertical. The ratio f/i - M is given a convenient value of, say, 100 (occa- sionally 50), whilst the additive constant (/ + d) = K will vary depend- ing upon the instrument. The formula may thus be simplified as D = M.s + K (4.29) Example 4.6 The constants M and K for a certain instrument were 100 and 1*5 respectively. Readings taken on to the vertical staff were 3*15, 4*26 and 5*37 ft respectively, the telescope being horizontal. Calculate the horizontal distance from the instrument to the staff. The stadia intercept s = 5-37-3-15 = 2-22 ft Horizontal distance D = 100 x 2-22 + 1-5 = 223-5 ft (68-1 m) If the instrument was set at 103-62 ft A.O.D. and the height to the trunnion axis at 4-83 ft, INSTRUMENTAL OPTICS 203 then the reduced level of the staff station = 103-62 + 4-83 - 4-26 = 104-19 ft A.O.D. (31-757 m) N.B. 4-26 - 3-15 = 5-37 - 4-26 = 1-11 = Is If the readings taken on to a metre staff were 0*960, 1*298, 1*636 respectively, then the horizontal distance = 100 x (1*636-0*960) = 67*6m + 0*5 m = 68*1 m If the instrument was set at 31*583 m A.O.D. and the height of the trunnion axis at 1*472 m, then the reduced level of the staff station = 31*583 + 1*472 - 1-298 = 31*757 m 4.56. The anallatic lens (Fig. 4.46) Vertical axis Anallatic Diaphragm ■•*>* Staff Fig. 4.46 The anallatic lens In the equation D = s(f/i) + (/ + d), the additive factor (/ + d) can be eliminated by introducing a convex lens between the objective and the diaphragm. The basic principles can be seen in Fig. 4.46. The rays from the staff Ad and Ce will for a given distance D always form a constant angle d intersecting at G. If this fixed point G is made to fall on the vertical axis of the instrument the additive term will be eliminated. Consider the object lens with the object AC and the image a, c,, i.e. neglecting the anallatic lens. By Eq.(4.19) and by Eq. (4.27) 1= 1+1 f U V u _ s v a,c. (4.30) (4.31) Consider the anallatic lens with the object as a^c x and the image 204 SURVEYING PROBLEMS AND SOLUTIONS as ac. Thus the object distance = v, - x and the image distance = v - x Applying the previous equations to this lens, i * _ * U ~ v i - * v - x (4.32) and v ' ~ * = J^L (4.33) v - x a x c t N.B. The object distance is assumed positive but the image distance is negative. An expression for D can now be found by eliminating v, v, and a^c^ from these four equations. From Eq. (4.32) v, - x = *< v ~ *> /, + v - * From Eq.(4.33) ac(v - x) a,Ci = v, - x Combining these gives a r - Q c (/i + v-x) "f 1 — 7 but from Eq. (4.30) v = -Hi Substituting in the above a i c t = but from Eq. (4.31) a,c, = giving sf u - f ac[f, + Uf - + u-f «) U sv u sf u - f ac(fi + uf - u-f ') i.e. s//, = aciu - O (/, + ■£- - *) Writing ac as i, the distance apart of the stadia lines sfU = HfM-f) + uf - *(«-/)] = *[«(/,+/-*) + /(*-/,)] u = s //i - /(* ~ /t) »(/+/,-*) / + /, - x INSTRUMENTAL OPTICS 205 but D = u + d t(/ + /,-*) f + U-x = Ms - ^ x ~ f'> + d and if d = ^-^ (4-34) f + U-x D = Ms (4.35) where M = //i i (/ + /,-*) (4.36) a constant factor usually 100. The manufacturer can therefore choose the lenses where the focal length / t is such that /, < x < f . Today, this is mainly of academic interest only, as all instruments have internal focussing telescopes, and the tacheometic formula D = /(s/i) + (f + d) is not applicable; nor can the internal focussing be considered anallatic as it is movable. The variation of the focal length of the objective system is gener- ally considered to be negligible for most practical purposes (see Ex- ample 4.7), manufacturers aiming at a low value for K, and in many cases the telescopes are so designed that when focussed at infinity the focussing lens is midway between the objective and the diaphragm. This allows accuracies for horizontal sights of up to 1/1000 for most distances required in this type of work. Example 4.7 An anallatic telescope is fitted with an object lens of 6 in. focal length. If the stadia lines are 0*06 in. apart and the vertical axis 4 in. from the object lens, calculate the focal length of the anal- latic lens and its position relative to the vertical axis if the multiplying constant is 100. From Eq. (4.34) the distance between objective and axis = 4 f+U-x When / = 6 in. d _ 6 (*-/i) 6 + /, - x Also, from Eq.(4.36), M ~ & !(/+/, -X) Therefore when M = 100, i = 0-06, /=6, M - 6/, 0-06(6 + U-x) 100 206 SURVEYING PROBLEMS AND SOLUTIONS Combining these equations, 4(6+/,-*) = 6(*-/,) 10* = 24 + 10/, and 100 x 0-06(6 + /,-*) = 6/, 6* = 36+0 * = 6 in. and /, = 3*6 in. Thus the focal length of the anallatic lens is 3*6 in. and its posi- tion is (6-4) = 2 in. from the vertical axis. Example 4.7a An anallatic tacheometer in use on a remote survey was damaged and it was decided to use a glass diaphragm not origin- ally designed for the instrument. The spacing of the outer lines of the new diaphragm was 0*05 in., focal lengths of the object glass and the anallatic lens 3 in., fixed distance between object glass and trunnion axis 3 in., and the anallatic lens could be moved by an adjusting screw between its limiting positions 3 in. and 4 in. from the object glass. In order to make the multiplier 100 it was decided to adjust the position of the anallatic lens, or if this proved inadequate to graduate a special staff for use with the instrument. Make calculations to de- termine which course was necessary, and if a special staff is required, determine the correct calibration and the additive constant (if any). What is the obvious disadvantage to the use of such a special staff? (L.U.) From Eq. (4.36), ,, M = ILL JL Mi / + u - u = 3 100 x 0-05 = 6 - 1-8 = 4-2 in. i.e. the anallatic lens should be 4*2 in. from the objective. As this is not possible, the lens is set as near as possible to this value, i.e. 4 in. Then M = ^-*- 3 - = 90 0-05(3 + 3 - 4) — The additive factor K from Eq.(4.34) = /(*-/,) = 3(4-3) = V5in /+/,-* 3 + 3-4 INSTRUMENTAL OPTICS 207 If the multiplying factor is to be 100, then the staff must be graduated in such a way that in reading 1 foot the actual length on the staff is 12 x 15 in. i.e. 13! in. 9 3 4.57. The tacheometric telescope (internal focussing) (Fig. 4.47) Mechanical axis Diaphragm H4 3£ ? u 2 Fig. 4.47 Tacheometric telescope (internal focussing) To find the spacing of the stadia lines to give a multiplying fac- tor M for a given sight distance: if i = stadia interval, and s = stadia intercept; for the convex lens (objective) s u, i.e. where m, is the magnifying power, for the concave lens (focussing) * _ v 2 = Vfir i.e. x = £Tl = m ,s i = xYl = m 2 x i = m^TTizS (4.37) but distance D = Ms - '-$ M (4.38) Example 4.8 An internally focussing telescope has an objective 6 in. from the diaphragm. The respective focal lengths of the objective and the internal focussing lens are 5 in and 10 in. Find the distance apart the stadia lines should be to have a multiplying factor of 100 for an observed distance of 500 ft. 208 SURVEYING PROBLEMS AND SOLUTIONS At 500 ft the object will be 500 ft - 6/2 in. from the objective, i.e. «, = 500 x 12 - 3 = 5997 in v i = mi X ? = 5-004 2 in. 5997 - 5 From Eq.(4.26), d 2 - d(/ + v,) + {v,(/ +/ 2 ) - / 2 /} = i.e. d z - d(6 + v,) + {16v, - 60} = d = | [(6 + v ) ± VK6 + v,) 2 - 64 v, + 240}] = |[(6 + v) + V(6-v 1 )(46-v 1 )]. i.e. d = i[ 11-004 2 ± VCO'995 8 x 40-995 8)] = 2-308 in. = / _ d = 6 - 2-308 = 3-692 in v u 2 = v,-d = 5-004 2-2-308 - 2-696 in From Eq.(4.38), . _ Dm } m 2 _ Dv^V2 M Mu x u 2 500xl2x 5-0042x3-692 100 x 5997 x 2-696 = 0-068 56 in. Example 4.9 What errors will be introduced if the previous instrument is used for distances varying from 50 to 500 ft ? At 50 ft rrt «, = 50 x 12 - 3 = 597 in. 597x5 = 2985 597 - 5 592 = 5-0422 in. Then, from Eq.(4.26), d = |[(6 + v,) + 4(6-^X46-^)] = |[ll-042 2 - VC0-9578 x 40-9578)] = 2-389 in. v 2 = 6 - 2-389 = 3-611 in. "2 = 5-042 - 2-389 = 2-653 in. INSTRUMENTAL OPTICS 209 The stadia intercept (s) = 0-068 56 x " 1 " 2 v,v 2 0-06856 x 597 x 2-653 5-042 x 3-611 = 5-9641 in. - 0-497 Oft The value should be 0-500 error = 0-0030 ft representing Q-30 ft in 100 ft At 100 ft error = 0-27 ft 200 ft error = 0-20 ft 300 ft error = 0-09 ft 400 ft error = 0-01 ft 500 ft error = 0-00 ft Example 4.10 An internal focussing telescope has an object glass of 8 in. focal length. The distance between the object glass and the diaphragm is 10 in. When the telescope is at infinity focus, the inter- nal focussing lens is exactly midway between the objective and the diaphragm. Determine the focal length of the focussing lens. At infinity focus the optical centre of the focussing lens lies on the line joining the optical centre of the objective and the cross-hairs, but deviates laterally 0*001 in. from it when the telescope is focussed at 25 ft. Calculate the angular error in seconds due to this cause. (L.U.) With the telescope focussed at infinity, v, = /, For the focussing lens, 1 1 1 u «2 V 2 = 1 1 l-d v, - d 1 1 U-d l-d = 1 8 -5 1 2 15 10-5 u = 7-5 in. Focal length of focussing lens With focus at 25 ft (assuming 25 ft from object lens.) u, = 25 x 12 = 300 v, = J£jL = 30 °x 8 = 8-2192 in. «, - /, 300 - 8 210 SURVEYING PROBLEMS AND SOLUTIONS From Eq. (4.26), d 2 - dO + Vt) + IV1O + /2) -/a/I i.e. d 2 - 18-219 2d + (143-836 - 75) = Solving for d, d = 5-348 /-if, \\i H-* oooi M T x ' x J X\ Fig. 4.48 With focus at 25 ft the image would appear at x, neglecting the internal focussing lens, i.e. OX = v,. With the focussing lens moving off line, the line of sight is now EXJ Z and all images produced by the objective appear as on this line. The line of sight through the objective is thus displaced XX, in the length v,. To calculate XX,. XX, XE /,/ 2 UE i.e. x = 0-001 x (/ - v,) l-d 0-001 x (10 - 8*219) 10 - 5-348 = 0-001781 = o-OOO 391 in. 4-552 To calculate the angular error (5), tan 8 = Ml OX _ X 8 = 206 265x0-000 391 9 . 8 nccondc Q.010 4.6 Instrumental Errors in the Theodolite 4.61 Eccentricity of the horizontal circle In Fig. 4.49, let 0, = vertical axis INSTRUMENTAL OPTICS 211 2 = Graduated circle axis 0^0 2 = e = eccentricity Z A, ~ 0,4 2 = r Fig. 4.49 If the graduated circle (Centre 2 ) is not concentric with the ver- tical axis (centre 0,) containing the readers A and B, the recorded value 6 will be in error by the angle a. As the instrument is rotated, the readers will successively occupy positions 4,6,, A Z B 2 , A 3 B 3 . a, = 0i-<£i tan- M E e sin <f> A Z E tan a, ~ r - e cos e sin</> (4.39) (4.40) Since e is small compared with r and as a is small, a rad ~ tan a Similarly, tan a 2 = e sin<f> a. r + e cos0 e sin0 (4.41) (4.40) If the readers are 180° apart, 4,0,8, is a straight line and the mean of the recorded values 6 give the true value of the angle <f>. 212 SURVEYING PROBLEMS AND SOLUTIONS i.e. <f> = 0, - a, = Z + a 2 •"' 2cf> = 0, + 2 as a, ~ a 2 4> = §(0, + 2 ) (4.42) N.B. (1) On the line X 2 , a = 0. (2) At 90° to this line, a = maximum. (3) If the instrument has only one reader, the angle should be repeated by transitting the telescope and rotating anticlockwise, thus giving recorded values 180° from original values This is of particular importance with glass arc theodolites in which the graduated circle is of small radius. To determine the amount of eccentricity and index error on the horizontal circle: (1) Set index A to 0° and read displacement of index B from 180°, i.e. 5,. (2) Set index B to 0° and read displacement of index A from 180°, i.e. 8 2 . (3) Repeat these operations at a constant interval around the plate, i.e. zeros at multiples of 10°. If the readers A and B are diametrically opposed, let 5, = dis- placement of reader B, from 180°, Fig. 4.50. Index A y at 0°. Index B, at 180° - 8,, i.e. 180 - (2a + A). Reading fli 180-6 Index error \ Fig. 4.50 Let § 2 = displacement of reader A z from 180°, Fig. 4.51, Index B 2 at 0. Index A 2 at 180 - 8 2 , i.e. 180 - (2a + A). INSTRUMENTAL OPTICS 213 Index error X Fig. 4.51 If there is no eccentricity and A and B are 180° apart, then 5, = S 2 = 0. If there is eccentricity and A and B are 180° apart, then 5, = 8 Z = a constant If there is no eccentricity and A and B are not 180° apart, then + 5, = -S 2 , i.e. equal, but opposite in sign. If there is eccentricity and A and B are not 180° apart, then 5, and 8 2 will vary in magnitude as the zero setting is consecutively changed around the circle of centre 2f but their difference will remain con- stant. A plotting of the values using a different zero for each pair of in- dex settings will give the results shown in Fig. 4.52. 360 +x and -x are 180" apart Fig. 4.52 4.62. The line of collimation not perpendicular to the trunnion axis Let the line of sight make an angle of 90° + € with the trunnion axis inclined at an angle a, Fig. 4.53. 214 SURVEYING PROBLEMS AND SOLUTIONS Vfcrtical axis Fig. 4. 53 Line of collimation not perpendicular to the trunnion axis The angular error Q in the horizontal plane due to the error e may be found by reference to Fig. 4.53. f = TY tan e YZ = TY cos a = tan € sec a tan0 = *I YZ tane = *X TY i.e. X] But TY = YZ sec a i.e. .-. tan0 = TYt * ne = TY cos a If and e are small, then 6 - = € sec a (4.43) (4.44) If observations are made on the same face to two stations of eleva- tions a , and a 2 , then the error in the horizontal angle will be ±(#i - 2 ) = ±tan -1 (tan £ sec a,) - tan"' (tan e sec a 2 ) (4.45) ±(0,-0 2 ) 2: ±e(seca, -seca 2 ) (4.46) On changing face, the error will be of equal value but opposite in sign. Thus the mean of face left and face right eliminates the error due to collimation in azimuth. The sign of the angle, i.e. elevation of de- pression, is ignored in the equation. The extension of a straight line, Fig. 4.54. If this instrument is used to extend a straight line by transitting the telescope, the follow- ing conditions prevail * With the axis on the line TQ the line of sight will be 0A V To observe A, the instrument must be rotated through the angle e to give pointing (1) — the axis will be rotated through the same angle E to T,Qr On transitting the telescope the line of sight will be (180° - 2a) A,OB, = A0B 2 . B 2 is thus fixed -pointing (2). On changing face the process is repeated — pointing (3) — and then INSTRUMENTAL OPTICS 215 Fig. 4.54 pointing (4) will give position S 4 . The angle B 2 0B A = 4e, but the mean position B will be the cor- rect extension of the line AO. The method of adjustment follows the above process, B 2 B 4 being measured on a horizontal scale. The collimation error may be corrected by moving the telescope graticule to read on B 3 , i.e. aB^. 4.63 The trunnion axis not perpendicular to the vertical axis (Fig. 4.55) The trunnion (horizontal or transit) axis should be at right angles to the vertical axis; if the plate bubbles are centralised, the trunnion axis will not be horizontal if a trunnion axis error occurs. Thus the line of sight, on transitting, will sweep out a plane inclined to the ver- tical by an angle equal to the tilt of the trunnion axis. Fig. 4.55 Trunnion axis not perpendicular to the vertical axis If the instrument is in correct adjustment, the line of sight sweeps out the vertical plane ABCD, Fig. 4.55. If the trunnion axis is tilted by an angle e, the line of sight sweeps 216 SURVEYING PROBLEMS AND SOLUTIONS out the inclined plane ABEF. In Fig. 4.55, the line of sight is assumed to be AE. To correct for the tilt of the plane it is necessary to rotate the horizontal bearing of the line of sight by an angle 6, to bring it back to its correct position. EC = BC tan e ED ED Thus sin# AD tane tana tane (4.47) (4.48) ED i.e. sin# = tana tane and if 6 and e are small, then 6 = e tan a where 6 = correction to the horizontal bearing e = trunnion axis error a = angle of inclination of sight On transitting the telescope, the inclination of the trunnion axis will be in the opposite direction but of equal magnitude. Thus the mean of face left and face right eliminates the error. Method of adjustment (Fig. 4.56) (1) Observe a highly elevated target A, e.g. a church spire. (2) With horizontal plates clamped, depress the telescope to observe a horizontal scale 8. (3) Change face and re-observe A. (4) As before, depress the telescope to observe the scale at C. (5) Rotate horizontally to D midway between B and C. (6) Elevate the telescope to the altitude of A. (7) Adjust the trunnion axis until A is observed. (8) On depressing the telescope, D should now be observed. Fig. 4.56 Field test of trunnion axis error INSTRUMENTAL OPTICS 217 4.64 Vertical axis not truly vertical (Fig. 4.57) If the instrument is in correct adjustment but the vertical axis is not truly vertical by an angle E, then the horizontal axis will not be truly horizontal by the same angle E. Thus the error in bearing due to this will be E tana (4.49) This is a variable error dependent on the direction of pointing relative to the direction of tilt of the vertical axis, and its effect is not eliminated on change of face, as the vertical axis does not change in position or inclination. Face left Horizontal line e 1/ t£^ i f / Face r ight Trunnion axis not perpendicular to the vertical axis Face left Horizontal lin e S \ V—. Fac e right Vertical axis unchanged on change of face Vertical axis not truly vertical Fig. 4.57 In Fig. 4.58, the true horizontal angle (0) A x OB y = angle (</>) A£)B 2 -( c i)Ei tana, + (c 2 ) E z tan a 2 . Thus the error in pointing (0) is dependent on (1) the tilt of the axis E, which itself is dependent on the direction of pointing, vary- ing from maximum (£ ) when on the line of tilt of the vertical axis to zero when at 90° to this line, and (2) the angle of inclination of the line of sight. To measure the value of e and E a striding level is used, Fig. 4.59. 218 SURVEYING PROBLEMS AND SOLUTIONS Fig. 4.58 Bubble axis Trunnion axis Vertical axis Face right Vertical axis Fig. 4.59 Let the vertical axis be inclined at an angle E Let the trunnion axis be inclined at an angle e Let the bubble be out of adjustment by an angle j3. Then face left tilt of the trunnion axis = E - e tilt of the bubble axis = E - j8 face right tilt of the trunnion axis = E + e tilt of the bubble a^is = E + jS INSTRUMENTAL OPTICS 219 Mean tilt of the trunnion axis = -|[(E - e) + (E + e)] = E (4.50) Mean tilt of the bubble axis = -±[(E - j8) + (E + j8)] = E (4.51) Therefore the mean correction taking all factors into account is Etana (4.52) N.B. The value of E is related to the direction of observation and its effective value will vary from maximum to nil. Tilting level readings should be taken for each pointing. If E is the maximum tilt of the axis in a given direction, then E, = E cos 6 where is the angle between pointing and direction of maximum tilt. Then the bubble recording the tilt does not strictly need to be in adjustment, nor is it necessary to change it end for end as some authors suggest, the mean of face left and face right giving the true value. If the striding level is graduated from the centre outwards for n pointings and 2n readings of the bubble, then the correction to the mean observed direction is given by c = A (XL -1R) tana (4.53) 2n where c = the correction in seconds d = the value of one division of the bubble in seconds XL = the sum of the readings of the left-hand end of the bubble 2/? = the sum of the readings of the right-hand end of the bubble a = the angle of inclination of sight n = the number of pointings. The sign of the correction is positive as stated. Any changes de- pends upon the sign of XL - 2/2 and that of a. N.B. The greater the change in the value of a the greater the effect on the horizontal angle. 4.65 Vertical circle index error (Fig. 4.60) When the telescope is horizontal, the altitude bubble should be central and the circle index reading zero (90° or 270° on whole circle reading instruments). If the true angle of altitude = a the recorded angles of altitude = a, and a 2 the vertical collimation error = <f> and the circle index error = 6, 220 SURVEYING PROBLEMS AND SOLUTIONS Recorded value (F.L.) = a = a, - <f> - 6 (F.R.) = a = a 2 + cf> + a = -1(0., + a 2 ) (4.54) Collimation error Bubble axis Eyepiece end Objective end Index error S Face left ^\ / Collimation ^X^" error & Ssf \ VC -''A «2 \ ^fe.~r* -. -4 Index error ble axis— { ^J Mf ^f^- \^s^ Face right Fig. 4.60 Thus, provided the altitude bubble is centralised for each reading, the mean of face left and face right will give the true angle of altitude. If the bubble is not centralised then bubble error will occur, and, depending on the recorded displacement of the bubble at the objective and eyepiece ends, the sensitivity will indicate the angular error. As the bubble is rotated, the index is also rotated. Thus 6 will be subjected to an error of ± i"(0 - E)8", where 8" = the angular sensitivity of the bubble. If the objective end of the bubble is higher than the eyepiece end on face left, i.e. L > E L , then 6 will be decreased by ±-(0 L - B L )8", i.e. F.L. a = a, - <f> - {6 - \{0 L -E L )8"\ and F.R. a = a 2 + cf> + {0 + Uo R - E R )8"\ INSTRUMENTAL OPTICS 221 a = |(a 1 + a 2 ) +y{(0 L+ /e ) ~(E L + E R )\ = i<a f + a 2 ) + |(20 - SB) (4.55) To *es* and adjust the index error (1) Centralise the altitude bubble and set the telescope to read zero (face left). (2) Observe a card on a vertical wall — record the line of sight at A. (3) Transit the telescope and repeat the operation. Record the line of sight at B. (4) Using the slow motion screw (vertical circle) observe the mid- point of AB. (The line of sight will now be horizontal.) (5) Bring the reading index to zero and then adjust the bubble to its midpoint. Example 4.11 Trunnion axis error. The following are the readings of the bubble ends A and B of a striding level which was placed on the trunnion axis of a theodolite and then reversed ('Left' indicates the left-hand side of the trunnion axis when looking along the tele- scope from the eyepiece end with the theodolite face right.) A on left 11-0, B on right 8*4 B on left 10*8, A on right 8*6 One division of the striding level corresponds to 15". All adjust- ments other than the horizontal trunnion axis adjustment of the theodo- lite being presumed correct, determine the true horizontal angle be- tween P and Q in the following observations (taken with the theodo- lite face left). Object Horizontal circle Vertical circle P 158° 20' 30" 42° 24' Q 218° 35' 42" 15° 42' (L.U.) By Eq.(4.53), Correction to bearing = -£-(%L - 2R) tana 2n to P c = ^{(11-0 + 10-8) - (8-4 + 8-6)}tan 42° 24' = 15 x 4 ' 8 tan 42° 24' 4 = -18" tan 42° 24" = -16" 222 SURVEYING PROBLEMS AND SOLUTIONS to Q c = -18 tan 15° 42' = - 5' N.B. The correction would normally be positive when using the gen- eral notation, but the face is changed by the definition given in the problem. True bearing to P = 158° 20' 30" - 16" = 158° 20' 14" True bearing to Q = 218° 35' 42"- 5" = 218° 35' 37" True horizontal angle = 60° 15' 23" Example 4.12 In an underground traverse the following mean values were recorded from station B on to stations A and C A C Striding level: Station Horizontal Vertical Observed Angle Angle 136° 21' 32" -13° 25' 20" +47° 36' 45" 1 division = 10 seconds bubble graduated to 20 Height of instrument at B 4-63 ft Height of target at A 3-42 ft at C 5-15 ft Ground length AB 256-32 ft BC 452-84 ft Calculate the gradient of the line AC Striding Level readings 17-4 14-5 5-8 2-7 (R.I.C.S.) Striding level corrections to A L 17-4 R 5-8 2 )23-2 11-6 i.e. centre of bubble is 1-6 to left of centre of graduations. to B L 14-5 R 2-7 2 ) 17-2 8-6 i.e. centre of bubble is 1-4 to right of centre of graduations. The same, results may be obtained by using the basic equation (4.53) and transposing the readings as though the graduation were from the centre of the bubble. INSTRUMENTAL OPTICS 223 i.e. to A 17-4 (L) becomes 7-4 (L) 5-8 (R) becomes 4*2 (R) XL - XR 7-4 - 4-2 = 1#6 2 2 to C 14-5 (L) becomes 4-5 (L) 2-7 (R) becomes 7-3 (R) XL - XR _ 4-5 - 7-3 = _ 1>4 2 ~ 2 Applying these values to Eq.(4.52), Correction to A = +1*6 x 10 x tan (-13° 25') = -3'8* Correction to C = -1-4 x 10 x tan(+47°37') = - 15,3 " Total angle correction = -11*5" Corrected horizontal angle = 136° 21' 32" - 11-5" = 136° 21' 20" To find true inclination of the ground and true distances (Fig. 4.61) Line AB Ba = sin" 1 1-21 cos 13° 25' 20" 1 256-32 5a, = -0°15'47" a, = 13° 25' 20" 6 = 13° 09' 33" Horizontal length (D,) AB = 256-32 cos 13° 09' 33" = 249-59 ft Vertical difference (tf,) AB = 256-32 sin 13° 09' 33" = 58-35 ft Line BC . 0-52 cos 47° 36' 45" 8a z = sin" 1 r=^rzrz 2 452-84 = -0°02'40" a 2 = 47° 36' 45" = 47° 34' 05" Horizontal length BC = 452-84 cos 47° 34' 05" = 305-54 ft Vertical difference = 452-84 sin 47° 34' 05" = 334-23 ft Difference in height AC = 58-35 + 334-23 = 392-58 ft To find the horizontal length AC : In triangle ABC, tan A -° 305-54 - 249-59 tan (180 - 136° 21' 20") 2 305-54 + 249-59 2 224 SURVEYING PROBLEMS AND SOLUTIONS 4-63 4-63 3-42 A TffKS^ 5-15 4-63 4-63 A-C A + C Fig. 4.61 2° 18' 40" 21°49'20' A = 24°08'00" Then AC = 305-54sinl36°21 , 20"cosec24°08 , 00" = 515-77 Gradient AC = 392*58 ft in 515'77ft = 1 in 1-314 INSTRUMENTAL OPTICS 225 Example 4.13 (a) Show that when a pointing is made to an object which has a vertical angle h with a theodolite having its trunnion axis inclined at a small angle i to the horizontal, the error introduced into the horizontal circle reading as a result of the trunnion axis tilt is i tan/i. (b) The observations set out below have been taken at a station P with a theodolite, both circles of which have two index marks. On face left, the vertical circle nominally records 90° minus the angle of elevation. The plate bubble is mounted parallel to the trunnion axis and is graduated with the zero of the scale at the centre of the tube one division represents 20". The intersection of the telescope cross-hairs was set on signals A and B on both faces of the theodolite. The means of the readings of the circle and the plate bubble readings were : Signal Face Horizontal Circle Vertical Circle Midpoint of Bubble A B The vertical axis was then rotated so that the horizontal circle reading with the telescope in the face left position was 256° 40'; the reading of the midpoint of the bubble was then 0*4 division away from the circle. If the effect of collimation error c on a horizontal circle reading is c sec/i, calculate the collimation error, the tilt of the trunnion axis and the index error of the theodolite, the altitude of the vertical axis when the above observations were taken, and the value of the horizon- tal angle APB. (N.U.) At A (Fig. 4.62). As the bubble reading is equal and opposite, on change of face the horizontal plate is horizontal at 90° to the line of sight. The bubble is out of adjustment by 1 division = -20" F.L. At 90° to A the corrected bubble reading gives F.L. 0-4 + 1-0 = +1-4 div. = +28" The horizontal plate is thus inclined at 28" as is the vertical axis, in the direction A. left 116° 39' 15" 90° 00' 15" 1-0 division towards circle right 346° 39' 29" 270° 00' 17" 1-0 division towards circle left 301° 18' 36" 80° 03' 52" central right 121° 18' 30" 279° 56' 38" 2-0 division towards circle 226 SURVEYING PROBLEMS AND SOLUTIONS Fig. 4.62 At B (Fig. 4.63). The corrected bubble readings give F. F. N.B. This value may be checked (see p. 414) \L. 0-0+ 1-0 = +1-CH \R 2-0-1-0 = +1-0 J i.e. +20' 166* (28") Fig. 4.63 INSTRUMENTAL OPTICS 227 Apparent dip = full dip cosine angle between = 28"cos(301-256) = 28" cos 45° = 20^_ The effect of instrumental errors in pointings (Fig. 4.64) Fig. 4.64 Horizontal Collimation 6 C = ±c sec/i (if h = Q c = c); the mean of faces left and right gives the correct value. Trunnion Axis $ t = ± i tan ft (if h = d t = 0); the mean of faces left and right give the correct value. Vertical Axis d v = v tan/i (if h = 6 V = 0); the sign is dependent on the inclination of the axis. (F.L. inclination towards circle, with +h, d v is -ve Vertical Index error Bh = 2\h t -h r \; the mean of faces left and right gives the correct value Application to given values: At A (F.L.) 166° 39' 15" + c sech - v tank + i tan/i i.e. 166° 39' 15" + c (F.R.) 346° 39' 29" - c F.L. must equal F.R. .-. 166° 39' 15" + c = 346° 39' 29" - 180° - c 2c = +14" c = + 7' (collimation error) 228 SURVEYING PROBLEMS AND SOLUTIONS At B (F.L.) 301° 18'36" + 7 sec9°57" - 20 tan9°57' + i tan 9° 57" i.e. 301°18'36" +7-1 - 3-5 + 0-175* (F.R.) 121°18'30" -7-1 - 3-5 - 0-175 i F.L. must equal F.R. /. 301°18'36" + 7-1 -3-5 + 0-175 i = 121° 18' 30" + 180 - 7-1 - 3-5 - -0-175 i i.e. 0-35 i = -20-2" i = -58" (fli =-10-1") (trunnion axis error) Corrected readings F.L. A 166° 39' 15" + 7" = 166° 39' 22" B 301° 18' 36" + 7-1" - 3-5" - 10-1 = 301° 18' 29-5" Angle APB = 134° 39' 07-5" F.R. A 346° 39' 29" - 7" = 346° 39' 22" B 121° 18' 30" - 7-1 - 3-5 + 10-1 = 121° 18' 29-5" Angle APB = 134° 39' 07-5" Vertical angles At A: 8h = I{(90-90°00'15") - (270°00'17" -270)} = -16" At B: Sh = |{(90-80°03'52") - (279° 56' 38" -270)} = -15" N.B. The discrepancy is assumed to be an observational error. 4.7 The Auxiliary Telescope This is used where steep sights are involved and in two possible forms: (1) Side telescope (2) Top telescope 4.71 Side telescope There are two methods of using this form of telescope: (a) in adjust- ment and (b) out of adjustment with the main telescope. Adjustment (a) Alignment (Fig. 4.65). Observe a point A with the main tele- scope. Turn in azimuth to observe with the side telescope without altering the vertical circle. Raise or lower the side telescope until the horizontal cross-hair coincides with the target A. The horizontal hairs are now in the same plane. INSTRUMENTAL OPTICS 229 Main telescope Side telescope Fig. 4.65 Alignment of telescopes Fig. 4.66 (b) Parallel lines of sight (Fig. 4.66). If x is the eccentricity of the telescope at the instrument this should be constant between the lines of sight (see Fig. 4.66). At a distance d x from the instrument, a scale set horizontally may be read as a,b, giving an intercept s,, and then at a\ readings a 2 b 2 give intercept s 2 . If the lines of sight are parallel, If not, the angle of convergence/divergence e is given as e = tan -i s 2 Si (4.56) d 2 — d, If s 2 > s,, the angle is +ve, i.e. diverging If s 2 < s,, the angle is -ve, i.e. converging. The amount of eccentricity x can be obtained from the same read- ings. 230 SURVEYING PROBLEMS AND SOLUTIONS d\ Si - X d 2 - d, i.e. x = Si(**2 - <*i) - ^1(^2 - Si) (d 2 -d,) s,d 2 — s 2 di (4.57) d 2 - d, By making the intercept s 2 = x, the collimation of the auxiliary telescope can be adjusted to give parallelism of the lines of sight. Observations with the side telescope (a) Vertical Angles. If the alignment is adjusted, then the true vertical angle will be observed. If an angular error of 8a exists between the main and the side telescope, then the mean of face left and face right observations is re- quired, i.e. F.L. = a, + da = a F.R. = a 2 - Sa = a a = i(a, + a 2 ) (4.58) (b) Horizontal Angles (Fig. 4.67) If Fig. 4.67 Horizontal angles with the side telescope 6 = the true horizontal angle; <f> = the recorded horizontal angle; S, and 8 2 = errors due to eccentricity, INSTRUMENTAL OPTICS 231 then = <t> x - 5, + 8 2 say F.L. 6 = <f> 2 + S, - 8 2 F.R. i.e. = ya(^, + <^) (4.59) Example 4.14 In testing the eccentricity of a side telescope, read- ings were taken on to levelling staves placed horizontally at X and Y 100 and 200 ft respectively from the instrument. Readings at X 5-42 ft 5-01 ft at Y 3-29 ft 2-79 ft Calculate (a) the collimation error (e), (b) the eccentricity (x). (R.I.C.S./M) From the readings, Si = 5-42 - 5-01 = 0-41 s 2 = 3-29 - 2-79 = 0-50 Then, by Eq. (4.56), e - fan"' S2 ~ Si d 2 - <*, €" 206265 x 200 (0-50 - 0-41) - 100 = 185-6" = 03' 06" by Eq. (4.57) X _ S\d 2 — s 2 d d 2 - d, i 0-41 x 200 - 0-50 x 100 200 - 100 82 - 50 0-32 ft Based on the metric system the question becomes: In testing the eccentricity of a side telescope, readings were taken on to levelling staves placed horizontally at X and Y, 30*48 m and 60*96 m respectively from the instrument. Readings at X 1*652 m 1*527 m at y 1*003 m 0*850 m Calculate (a) the collimation error (€), (b) the eccentricity (x) s, = 1*652 - 1*527 = 0*125 m s 2 = 1*003 - 0*850 = 0*153 m Then e = 206 265 x (0*153-0*125) 60*96 - 30*48 = 189*5" = 03' 10" 232 and SURVEYING PROBLEMS AND SOLUTIONS 0-125 X 60-96 - 0-153 x 30-48 X ~~ 30-48 = 0-097 m (0-32 ft) The effect of eccentricity x and collimation error e In Fig. 4.68, assuming small 4 angles, Angle AOB = 8" 206 265 x Angle BSC = e" 206 265 y (4.60) (4.61) Angle AOC 206265(x+y) (4 62) d Fig. 4.68 e = 8 + e (4.63) ••• Angle BOC = Angle BSC = £ As the eccentricity x is con- stant, the angle (5) is dependent upon the length of sight d. As the collimation angle € is constant, it has the same effect as the collimation error in the main telescope. It affects the horizontal angle by e sec a, where a is the vertical angle. Assuming the targets are at different altitudes, the true horizontal angle #,Fig. 4.67, is given as 6 = <£, - (5, + €seca,) + (5 2 + eseca 2 ) say F.L. (4.64) Also 6 = <j> 2 + (8 2 + eseca,) - (5 2 + e seca 2 ) F.R. (4.65) 6 = %(<£, + 2 ) (4.66) Thus the mean of the face left and face right values eliminates errors from all the above sources. Example 4.15 Using the instrument of Example 4.14, the following data were recorded: INSTRUMENTAL OPTICS (€ = 03' 06" x = 0-32 ft) 233 Station set Station Horizontal Vertical Remarks at observed circle circle B A 0°05'20" + 30° 26' Horizontal lengths C 124° 10' 40" -10° 14' AB = 100' BC = 300' side telescope on right. Calculate the true horizontal angle ABC (R.I.C.S./M) By Eq. (4.64), True horizontal angle (6) = <f> - (5,+ e sec a,) + (5 2 + e secci^ ) 206265 x 0-32 = 66 » 5, = 6\ = 100 206265 x 0-32 300 1'06' = 22" = 0'22" eseca, = 186" sec30°26' = 215-7", say 216" = 3' 36" eseca 2 = 186" sec 10° 14' = 189*0" = 3' 09" <f> = 124° 10' 40" - 0° 05' 20" = 124°05'20" 6 = 124°05'20" - (1'06"+ 3'36") + (0' 22"+ 3' 09") = 124° 04' 09" 4.72 Top telescope In this position the instrument can be used to measure horizontal angles only if it is in correct adjustment, as it is not possible to change face. Adjustment (a) Alignment. The adjustment is similar to that of the side tele- scope but observations are required by both telescopes on to a plumb line to ensure that the cross-hairs are in the same plane. (b) Parallel lines of sight (Fig. 4.69). Here readings are taken on vertical staves with the vertical circle reading zero. The calculations are the same as for the side telescope: 6' = tan-' S z ~ s ' d, - d, (4.67) and Sid 2 — s 2 d, d 2 ~ dy (4.68) 234 SURVEYING PROBLEMS AND SOLUTIONS £' . " Si — -~. V • / <*1 <*2 Fig. 4.69 Measurement of Vertical Angles (Fig. 4.70) Angle ATB is assumed equal to angle AOB as both are small values Fig. 4.70 Measurement of vertical angles with the top telescope The angular error (§') due to eccentricity is given as 8' = tan-' (4.69) where d sec0 x = eccentricity d = horizontal distance <£ = recorded vertical angle If e' is the error due to collimation then the true vertical angle (0) is given as ±6 = ±(0+S'+ €') (4.70) assuming 8' and e' small. 4.8 Angular Error due to Defective Centring of the Theodolite The angular error depends on the following, Fig. 4.72: (a) linear displacement x, (b) direction of the instrument B, relative to the station B (c) length of lines a and c. INSTRUMENTAL OPTICS 235 Locus of centring error Fig. 4.71 Minimum error due to defective centring of the theodolite The instrument may be set on the circumference of the circle of radius x. No error will occur if the instrument is set up at B, or B 2 (Fig. 4.71), where A,B y ,B,B 2 and C lie on the arc of a circle. Fig. 4.72 The effects of centring errors In Fig. 4.72, let the instrument be set at B t instead of B. Angle 0, is measured instead of 6 i.e. 6 = 0, - (a + B) Assume the misplumbing x to be in a direction <f> relative to the line AB. In triangle ABB U sin a = x sin<£ AB, (4.71) As the angle a is small, a _ 206265 x sin <f> c (4.72) Similarly, B" = 206265 x sin (0 - <^) (4 73) a .'. Total error E = a + B = 206265 Je^ + sin(fl - <ft) 1 (4.74) For maximum and minimum values, 4*L = 206265 r[ cos< £ _ cos(fl - 0) 1 = Q 236 SURVEYING PROBLEMS AND SOLUTIONS COS 4> _ COS (0 - (f>) i.e. c cos <ji = -£(cos 6 cos <£ + sin 6 sin </>) -j-sin<£ cot0 = £(cos0 cot<£ + sin0) cot0(l- ccos ^ = a c sin 6 cot<£ = csind n (4.75) a - c cos d N.B. (1) If = 90°, 6 = or 180° (2) If a :» c, then <f> -* 90°, i.e. the maximum error exists when (f> tends towards 90° relative to the shorter line. (3) If a = c, <f> = 6/2. Professor Briggs proves that the probable error in the measured angle is , . ±2» //J: + I _!«££) (4.76) Example 4. 16 The centring error in setting up the theodolite at station B in an underground traverse survey is ±0*2 in. Compute the maximum and minimum errors in the measurement of the clockwise angle ABC induced by the centring error if the magnitude of the angle is approximately 120° and the length of the lines AB and BC is ap- proximately 80*1 and 79*8 ft respectively. (R.I.C.S.) (1) The minimum error as before will be nil. (2) The maximum error on the bisection of the angle ABC is AB ~ BC. i.e. <f> = = 60° a ~ c ~ 80 ft x = 9^1 = 0-0167 ft 12 E = 206265 x 0-016 7[ sin60 + sin(120-60) l L 80 80 J = 206265 x 0-0167 x 2 sin 60/80 = 74 seconds i.e. l' 14" By Professor Briggs' equation, the probable error e = ± 2 x0-016 7 //_1_ J_ _ 2 cos 120 \ 3-1416 aJ\S0 2 + 80 2 80+80 / = ±35" i.e. a 1/2 max error. INSTRUMENTAL OPTICS 237 4.9 The Vernier This device for determining the decimal parts of a graduated scale may be of two types: (1) Direct reading (2) Retrograde both of which may be single or double. 9 l (a) (b) | \ ■■■■■ *' | ' ■' ■ ' ■ ' I | . . . i | i i i i ) i i 10 20 ® 5 ■i I i — i i i | — i v I j '| ' ' ' ' ' ' ' ' i » — i — i i I — r ^ 30 \ | 40 50 Vernier scale 36-0 ■*• 0-3 =» 36-3 reading I 5 © ' ■'■ ' ■ | . '■ X \ N ■■ ■ ■ ■ ■ i ■ ■ ' ■ r ■ ■ ■ ■ i ' • i {C) 30 3^.3 40 50 Retrograde vernier Fig. 4.73 Verniers 4.91 Direct reading vernier Let d = the smallest value on the main scale v = the smallest value on the vernier scale n = number of spaces on the vernier n vernier spaces occupy (ji - 1) main scale spaces i.e. nv = (n - l)d v = (n-l)d n Therefore the least count of the reading system is given by: d-v - d - «" - » = d d - v = d( n - n + *) = I \ n / n (4.77) (4.78) Thus the vernier enables the main scale to be read to -th of 1 division. 238 SURVEYING PROBLEMS AND SOLUTIONS ii Example 4. 17 If the main scale value d = -^ and the number of spaces on the vernier (n) = 10, the vernier will read to 1/10 x 1/10 = 1/100 in. 4.92 Retrograde vernier In this type, n vernier division occupy (n + 1) main scale divi- sions, i.e. nv = (n + l)d -'fcr 1 ) v = d HJ^ (4.79) The least count = v - d = *m teM v - d = — as before (4.80) n 4.93 Special forms used in vernier theodolites In order to provide a better break down of the graduations, the vernier may be extended in such a way that n vernier spaces occupy (mn - 1) spaces on the main scale, (m is frequently 2.) nv = (jnn - l)d The least count = md - v = md . d(HLzl) (4.81) = d md - v = — as before (4.82) 4.94 Geometrical construction of the vernier scale In Fig. 4.74(a) the main scale and vernier zeros are coincident. For the direct reading vernier 10 divisions on the vernier must occupy 9 divisions on the main scale. Therefore (1) Set off a random line OR of 10 units. (2) Join R to V i.e. the end of the random line R to the end of the vernier V. (3) Parallel through each of the graduated lines or the random line INSTRUMENTAL OPTICS 239 to cut the main scale so that 1 division of the vernier = 0*9 divisions of the main scale. 10 R Vernier scale 36-0 ♦ 0-3= 36-3 reading Fig. 4.74 Construction of a direct reading vernier To construct a vernier to a given reading In Fig. 4.74(b) the vernier is required to read 36*3. It is thus re- quired to coincide at the 3rd division, i.e. 3 x 0*9 = 2*7 main scale division beyond the vernier index. Therefore coincidence will occur at (36*3 + 2*7) = 39*0 on the main scale and 3 on the vernier scale. The vernier is constructed as above in the vicinity of the point of coincidence. The appropriate vernier coincidence line (i.e. 3rd) is joined to the main scale coincidence line (i.e. 3S9*0) and lines drawn parallel as before will produce the appropriate position of the vernier on the main scale. In the case of the retrograde vernier, Fig. 4.75, 10 divisions on the vernier equals 11 divisions on the main scale, and therefore the point of coincidence of 3 on the vernier with the main scale value is 36-3 - (3 x 1-1) = 36-3 - 3-3 = 33-0 i i i I i i i i i i i i i 30 | 40 50 Coincidence Fig. 4. 75 Construction of a retrograde vernier 240 SURVEYING PROBLEMS AND SOLUTIONS Example 4.18 Show how to construct the following verniers: (1) To read to 10" on a limb divided to 10 minutes. (2) To read to 20" on a limb divided to 15 minutes. (3) The arc of a sextant is divided to 10 minutes. If 119 of these divisions are taken as the length of the vernier, into how many divi- sions must the vernier be divided in order to read to (a) 5 seconds (b) 10 seconds? (ICE.) (1) The least count of the vernier is given by Eq. (4.77) as d/n 10" - 10 x 60 ~ n n = = 60 10 — Therefore the number of spaces on the vernier is 60 and the number of spaces on the main scale is 59. (2) Similarly, 20" = 11 60 n ... „ = llx«! = 45 20 — i.e. the number of spaces in the vernier is 45 and the number of spaces on the main scale is 44. (3) The number of divisions 119 is not required, and the calculation is exactly as above. (a) n = 10 * 60 = 120 (b) „ , 10 *60 , 60 10 — Exercises 4(b) 3. The eccentricity of the line of collimation of a theodolite telescope in relation to the azimuth axis is 1/40 th of an inch. What will be the difference, attributable to this defect, between face right and face left measurement of an angle if the lengths of the drafts adjacent to the in- strument are 20 ft and 120 ft respectively ? (M.Q.B./S Ans. 17-9") 4. A horizontal angle is to be measured having one sight elevated to 32° 15 whilst the other is horizontal. If the vertical axis is inclined at 40 to the vertical, what will be the error in the recorded value ? (Ans. 25") 5. In the measurement of a horizontal angle the mean angle of eleva- tion of the backsight is 22° 12' whilst the foresight is a depression INSTRUMENTAL OPTICS 241 of 37° 10'. If the lack of vertically of the vertical axis causes the horizontal axis to be inclined at 50" and 40" respectively in the same direction, what will be the error in the recorded value of the horizontal angle as the mean of face left and right observations ? (Ans. 51") 6. In a theodolite telescope the line of sight is not perpendicular to the horizontal axis in but in error by 5 minutes. In measuring a horizontal angle on one face, the backsight is elevated at 33° 34' whilst the fore- sight is horizontal. What error is recorded in the measured angle ? (Ans. 60") 7. The instrument above is used for producing a level line AB by transiting the telescope, setting out C,, and then by changing face the whole operation is repeated to give C 2 . If the foresight distance is 100 ft, what will be the distance between the face left and face right positions, i.e. C,C 2 ? (Ans. 6-98 in.) 8. Describe with the aid of a sketch, the function of an internal focussing lens in a surveyors telescope and state the advantages and disadvantages of internal focussing as compared to external focussing. In a telescope, the object glass of focal length 6 in. is located 8 in. from the diaphragm. The focussing lens is midway between these when a staff 80 ft away is focussed. Determine the focal length of the focussing lens. (L.U. Ans. 4-154 in.) 9. In testing the trunnion axis of a vernier theodolite, the instrument was set up at '0', 100 ft from the base of the vertical wall of a tall building where a well-defined point A was observed on face left at a vertical angle of 36° 52'. On lowering the telescope horizontally with the horizontal plate clamped, a mark was placed at B on the wall. On changing face, the whole operation was repeated and a second position C was fixed. If the distance BC measured 0*145 ft, calculate the inclination of the trunnion axis. , „ (Ans. 3' 20 ) 10. The following readings were taken on fine sighting marks at B and C from a theodolite station A. Instrument Tq Vertical F.R.S.R. F.L.S.L. At Angle Vernier A Vernier B Vernier A Vernier B A B 72° 30 ' 292°26'30" 112°26'30" 23° 36' 24" 203° 36' 24" C -10° 24' 52° 39' 36" 232° 39' 36" 143°50'lo" 323° 50* 10" B 72° 30 ' 292°26'30" 112° 26*30" 23° 36' 24" 203° 36* 24" Calculate the value of the horizontal collimation error assuming 242 SURVEYING PROBLEMS AND SOLUTIONS this to be the only error in the theodolite and state whether it is to the right or left of the line perpendicular to the trunnion axis when the in- strument is face left. Describe as briefly as possible how you would adjust the theodo- lite to eliminate this error. , . n _ .„ . , v (L.U. Ans. 8-66 left) 11. The focal length of object glass and anallatic lens are 5 in and 4jin. respectively. The stadia interval was 0*1 in. A field test with vertical staffing yielded the following: Inst Staff Staff Vertical Measured Horizontal Station Station Intercept Angle Distance (ft) P Q 2-30 +7° 24' 224-7 R 6-11 -4° 42' 602-3 Find the distance between the object glass and anallatic lens. How far, and in what direction, must the latter be moved so that the multiplying constant of the instrument is to be 100 exactly ? (L.U. Ans. 7-25 in.; 0*02 in. away from objective) 12. An object is 20 ft from a convex lens of focal length 6 in. On the far side of this lens a concave lens of focal length 3 in. is placed. Their principal axes are on the line of the object, and 3 in. apart. De- termine the position, magnification and nature of the image formed. (Ans. Virtual image 43 in. away from the concave lens towards object; magnification 0-28) 13. A compound lens consists of two thin lenses, one convex, the other concave, each of focal length 6 in. and placed 3 in apart with their principal axes common. Find the position of the principal focus of the combination when the light is incident first on (a) the convex lens and (b) the concave lens. (Ans. (a) real image 6 in. from concave lens away from object, (b) real image 18 in. from convex lens away from object) 14. Construct accurately a 30 second vernier showing a reading of 124° 23 30 ' on a main scale divided to 20 minutes. A straight line may be used to represent a sufficient length of the arc to a scale of 0*1 in. to 20 min. (N.R.C.T.) 15. (a) Explain the function of a vernier. (b) Construct a vernier reading 4*57 in. on a main scale divided to 1/10 in. (c) A theodolite is fitted with a vernier in which 30 vernier divi- sions are equal to 14° 30' on a main scale divided to 30 minutes. Is the vernier direct or retrograde, and what is its least count ? (N.R.T.C. Ans. direct; 1 min.) INSTRUMENTAL OPTICS 243 Bibliography THOMAS, W.N. , Surveying (Edward Arnold) SANDOVER, J. A., Plane Surveying (Edward Arnold) GLENDENNING, J., Principles and Use of Surveying Instruments (Blackie) CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) HIGGINS, A.L., Elementary Surveying (Longmans) TAYLOR, W.E., Theodolite Design and Construction (Cooke, Troughton & Simms) CURTIN, w. and LANE, R.F., Concise Practical Surveying (English Universities Press) SHEPPARD, J.S. Theodolite Errors and Adjustments (Stanley) The Theodolite and Its Application English version by A.H. Ward, F.R.I.C.S. (Wild (Heerbrugg)) JAMESON, A.H. Advanced Surveying (Pitman) higgins, A.L. Higher Surveying (Macmillan) SMIRNOFF, M.V., Measurements for Engineering and Other Surveys (Prentice -Hall) BANNISTER, A. and RAYMOND, s., Surveying (Pitman) DIBDIN, F.S.H., Essentials of Light (Macmillan) NELKON, M., Light and Sound (Heinemann) STANLEY, W.W., Introduction to Mine Surveying (Stanford University Press) LEVELLING 5.1 Definitions Levelling is the process concerned with the determination of the differences in elevation of two or more points between each other or relative to some given datum. A Datum may be purely arbitrary but for many purposes it is taken as the mean sea level (M.S.L.) or Ordnance Datum (O.D.). A Level Surface can be defined as a plane, tangential to the earth's surface at any given point. The plane is assumed to be perpen- dicular to the direction of gravity which for most practical purposes is taken as the direction assumed by a plumb-bob. A level line, Fig. 5.1, is a line on which all points are equidistant from the centre of gravity. Therefore, it is curved and (assuming the earth to be a sphere) it is circular. For more precise determinations the geoidal shape of the earth must be taken into consideration. Horizontal line Vertical Fig. 5.1 A Horizontal Line, Fig. 5.1, is tangential to a level line and is taken, neglecting refraction, as the line of collimation of a perfectly adjusted levelling instrument. (As the lengths of sights in levelling are usually less than 450 ft, level and horizontal lines are assumed to be the same— see §5.6 Curvature and Refraction.) The Line of Collimation is the imaginary line joining the inter- section of the main lines of the diaphragm to the optical centre of the object-glass. Mean Sea Level. This is the level datum line taken as the LEVELLING 245 reference plane. In the British Isles the Ordnance Survey originally accepted the derived mean sea level value for Liverpool. This has been superseded by a value based upon Newly n in Cornwall. Bench Mark (7\) (B.M.). This is a mark fixed by the Ordnance Survey and cut in stable constructions such as houses or walls. The reduced level of the horizontal bar of the mark is recorded on O.S. maps and plans. Temporary Bench Mark (T.B.M.). Any mark fixed by the observer for reference purposes. Backsight (B.S.) is the first sight taken after the setting up of the instrument. Initially it is usually made to some form of bench mark. Foresight (F.S.) is the last sight taken before moving the instrument. Intermediate Sight (I.S.) is any other sight taken. N.B. During the process of levelling the instrument and staff are never moved together, i.e. whilst the instrument is set the staff may be moved, but when the observations at one setting are completed the staff is held at a selected stable point and the instrument is moved forward. The staff station here is known as a Change Point (C.P.). 5.2 Principles Let the staff readings be a, b, c etc. Change I.S. point (CR) Inverted staff reading (f) F.S. A B C D Fig. 5.2 £ F In Fig. 5.2, Difference in level A to B = a-b a <b •'. -ve i.e. fall B to C = b-c b> c •'• +ve i.e. rise C to D = c-d c <d ••• -ve i.e. fall DtoE = e-(-/) inverted staff •'• +ve i.e. rise Eto F = -f-g -ve i.e. fall AtoF = (a-fo)+ (b- = a - d + e - -c) g + (c-d) + <«+/) + <-/-*) 246 SURVEYING PROBLEMS AND SOLUTIONS = (a+e)- (d+g) = 2B.S. -2 F.S. 2 rises = (b-c) + (e+f) 2 falls = (b-a)+(d-c)+(f+g) 2 rises - 2 falls = (a + e) -(d + g) = 2B.S. - 2F.S. (5.1) The difference in level between the start and finish = 2B.S. -2F.S. = 2 rises - 2 falls (5.2) N.B. (1) Intermediate values have no effect on the final results, and thus reading errors at intermediate points are not shown up. (2) Where the staff is inverted, the readings are treated as nega- tive values and indicated in booking by a bracket or an asterisk. 5.3 Booking of Readings 5.31 Method 1, rise and fall B.S. I.S. F.S. Rise Fall Reduced Level a b b —a X x-(b-a) c b-c x- (b-a)+ (b-c) e d d-c x - (b-a)+ (b-c)- (d-c) [/] «-(-/) x-(b-a)+(b-c)-(d-c) + (e + f) g f+g x-(b-a) + (b-c)- (d-c) + (e+f)- (f+g) a + e - (d + g) d + g (b-c)+(e+f) (b-a)+ (d-c) + (/+*) Example 5.1 Given the following readings: a = 2-06 e = 7'41 b = 5-13 / = -6-84 c = 3-28 g = 3-25 d = 3-97 N.B. (1) The difference between adjacent readings from the same in- strument position gives rise or fall according to the sign + or -. (2) At the change point B.S. and F.S. are recorded on the same line. LEVELLING 247 (3) Check 2 B.S. - 2 F.S. = 2 Rise - 2 Fall before working out reduced levels. Difference between reduced levels at start and finish must equal 2 B.S. - 2 F.S. B.S. I.S. F.S. Rise Fall Reduced Level Remarks 2-06 100-00 St. A. B.M. 100-00 A.O.D. 5-13 3-07 96-93 St. B. 3-28 1-85 98-78 St. C. 7-41 3-97 0-69 98-09 St. D. C.P. [6-84] 14-25 112-34 St. E. Inverted staff on girder 3-25 7-22 10-09 102-25 St. F. 9-47 16-10 13-85 102-25 -7-22 -13-85 100-00 + 2-25 + 2-25 + 2-25 5.32 Method 2, height of collimation Height of Collimation B.S. I.S. F.S. a b c e [/] d i a + e b + c-f d + g x + a x+a—d+e Reduced Level x x + a-b x + a — c x + a-d x + a-d + e-(-f) x+a-d+e-g 6x+5a-b-c-3d + 2e+f-g Arithmetical Check 2 Height of each collimation x no. of applications = 3(x + a) + 2(x + a-d + e) = 5x + 5a - 2d + 2e 2 Reduced levels - first 2 I.S. 2 F.S. = 5x + 5a-b-c-3d + 2e+f-g + b + c - f + d + g 5x + 5a -2d + 2e Thus the full arithmetical check is given as: 2 Reduced levels less the first + 2 I.S. + 2 F.S. should equal 2 Height of each collimation x no. of applications (5.3) Using the values in Example 5.1, 248 SURVEYING PROBLEMS AND SOLUTIONS B.S. I.S. F.S. Height of Collimation Reduced Level Remarks 2-06 102-06 100-00 St. A B.M. + 100-00 A.O.D. 5-13 96-93 St. B 3-28 98-78 St. c 7-41 3-97 105-50 98-09 St. D C.P. [6-84] 112-34 St. E Inverted staff on girder 3-25 102-25 508-39 St. F 9-47 8-41 7-22 7-22 -6-84 2-25 1-57 Check 508*39 + 1-57 + 7-22 517-18 102-06x3 = 306-18 105-50x2 = 211-00 = 517-18 N.B. (1) The height of collimation = the reduced level of the B.S. + the B.S. reading. (2) The reduced level of any station = height of collimation - reading at that station. (3) Whilst 2B.S.- 2 F.S. = the difference in the reduced level of start and finish this does not give a complete check on the intermediate values; an arithmetical error can be made without being noticed. (4) The full arithmetical check is needed to ensure there is no arithmetical error. On the metric system the bookings would appear thus: (for less accurate work the third decimal place may be omitted) B.S. I.S. F.S. Rise Fall Height of Collimation Reduced Level Remarks 0-628 2-259 1-564 1-000 [2-085] 1-210 0-991 0-564 4-344 0-936 0-210 3-076 31-108 32-157 30-480 29-544 30-108 29-898 34-242 31-166 St. A 30-480m A.O.D. 2-887 2-201 2-564 2-085 0-479 2-201 4-908 4-222 4-222 154-958 0-686 0-686 0-686 Check on collimation 154-958 + 0-479 + 2-201 = 157-638 LEVELLING 249 31-108 x 3 = 93-324 32-157 x 2 = 64-314 157-638 Example 5.2 Using the height of collimation calculate the respective levels of floor and roof at each staff station relative to the floor level at A which is 20 ft above an assumed datum. It is important that a complete arithmetical check on the results should be shown. Note that the staff readings enclosed by brackets thus (3*43) were taken with the staff reversed. B.S. I.S. F.S. Height of Collimation Reduced Level Horizontal Distance (ft) Remarks 2-47 22-47 20-00 Floor at A (3-43) 25-90 Roof at A 3-96 18-51 50 Floor at B (2-07) 24-54 50 Roof at B 4-17 18-30 100 Floor at C (1-22) 23-69 100 Roof at C 3-54 18-93 150 Floor at D (4-11) (2-73) 21-09 25-20 150 Roof at D 1-96 19-13 200 Floor at E (5-31) 26-40 200 Roof at E 2-85 18-24 250 Floor at F (3-09) 24-18 250 Roof at F 4-58 16-51 300 Floor at G (3-56) (1-16) 18-69 22-25 300 Roof at G 2-22 16-47 350 Floor at H (4-67) 23-36 350 Roof at H 1-15 17-54 400 Floor at / (6-07) 24-76 400 Roof at / + 2-47 + 24-43 -9-96 363-91 -7-67 -19-79 -5-20 + 4-64 -(-9-96) 4-76 + 4-76 Checks 1. ZB.S. -2F.S. = 4-76 = diff. in level A-I 2. (a) 2 Reduced levels except first 363*91 (b) 22-47 x 7 = 157-29 21-09 x 6 = 126-54 18-69 x 4 = 74-76 358-59 250 SURVEYING PROBLEMS AND SOLUTIONS (c) 2F.S. + 2I.S. = -9-96 + 4-64 -5-32 5-32 363*91 Checks with (a) N.B. Inverted staff readings must always be treated as negative values. Example 5.3 The following readings were taken with a level and a 14 ft staff. Draw up a level book page and reduce the levels by (a) the rise and fall method, (b) the height of collimation method. 2-24, 3-64, 6-03, 11-15, (12'72 and 1-48) C.P., 4'61, 6*22, 8'78, B.M. (102-12 ft A.O.D.), 11-41, (13'25 and 6'02) C.P., 2-13, 5-60, 12-21. What error would occur in the final level if the staff had been wrongly extended and a plain gap of 0*04 had occurred at the 5 ft sec- tion joint? (L.U.) B.S. I.S. F.S. Rise Fall Height of Collimation Reduced Level Remarks 2-24 3-64 6-03 11-15 1-40 2-39 5-12 122-14 119-90 118-50 116-11 110-99 1-48 4-61 6-22 12-72 1-57 3- 13 1-61 110-90 109-42 106-29 104-68 8-78 2-56 102-12 B.M. 102-12 A.O.D. 11-41 2-63 99-49 6-02 2-13 5-60 13-25 12-21 3-89 1-84 3-47 6-61 103-67 97-65 101-54 98-07 91-46 9-74 38-18 3-89 32-23 28-44 38-18 32-23 28-44 28-44 A combined booking is shown for convenience. If a 0*04 ft gap occurred at the 5 ft section all readings > 5 ft will be 0-04 ft too small. LEVELLING 251 The final level value will only be affected by the B.S. and F.S. readings after the reduced level of the datum, i.e. 102*12, although the I.S. 8*78 would need to be treated for booking purposes as a B.S. i.e. 1 B.S. = 8-78+ 6*02 = 14*80 2 F.S. - 13-25 + 12-21 = 25-46 Difference = - 10*66 .*. Final level = 102' 12 - 10*66 = 91*46 As all these values are > 5 ft no final error will be created, since all the readings are subjected to equal error. Therefore the final reduced level is correct, i.e. 91*46 ft A.O.D. Example 5.4 The following readings were observed with a level: 3*75 (B.M. 112*28), 5*79, 8*42, 12*53, C.P., 4*56, 7*42, 2*18, 1*48, C.P., 12*21, 9*47, 5*31, 2*02, T.B.M. (a) Reduce the levels by the Rise and Fall method. (b) Calculate the level of the T.B.M. if the line of collimation was tilted upwards at an angle of 6min and each backsight length was 300ft and the foresight length 100 ft. (c) Calculate the level of the T.B.M. if the staff was not held up- right but leaning backwards at 5° to the vertical in all cases. (L.U.) (a) B.S. I.S. F.S. Rise Fall Reduced Level Remarks 3-75 5-79 8-42 2-04 2-63 112-28 110-24 107-61 B.M. 112-28 4-56 7-42 2-18 12-53 5-24 4-11 2-86 103-50 100-64 105-88 12-21 9-47 5-31 1-42 0-76 2-74 4-16 106-64 109-38 113-54 2-02 3-29 116-83 T.B.M. 20-52 15-97 16-19 11-64 15-97 11-64 4-55 4-55 4-55 252 SURVEYING PROBLEMS AND SOLUTIONS 39 300 ft 100ft Fig. 5.3 (b) In Fig. 5.3, True difference in level = (a-3e) - (b-e) = (a-b) - 2e where e = 100 tan 06' = 100 x 06' radian8 = 1°° x 6 x 60 = 0-175 per 100 ft 206 265 ~ Total length of backsight = 3 x 300 = 900 ft of foresight = 3 x 100 = 300 ft Effective difference in length = 600 ft .". Error = 6 x 0*175 = 1*050 ft i.e. B.S. readings are effectively too large by 1*05 ft. /. True difference in level = 4*55 - 1*05 = 3*50 ft .'. Level of T.B.M. = 112*28 + 3*50 = 115*78 ft A.O.D. (c) If the staff was not held vertical the readings would be too large, the value depending on the staff reading. True reading = observed reading x cos 5° Line of sight Apparent difference in level 2 B.S. - 2F.S. = 4*55 True difference in level = 2 B.S. cos 5° - 2 F.S. cos 5° = (2 B.S. -2 F.S.) cos 5° Fig. 5.4 = 4*55 cos 5° = 4*53 Level of T.B.M. = 112*28 + 4*53 = 116*81 ft Example 5.5 Missing values in booking. It has been found necessary to consult the notes of a dumpy levelling carried out some years ago. LEVELLING 253 Whilst various staff readings, rises and falls and reduced levels are undecipherable, sufficient data remain from which all the missing values can be calculated. B.S. I.S. F.S. Rise Fall Level A.O.D. Distance Remarks 2.36 1-94 121.36 100 B.M. on Watson House 4-05 4-31 6.93 7-29 4.46 113-32 200 300 400 500 600 7-79 0-63 715 Peg 36 3-22 1-58 112-01 800 900 6-53 5-86 113-53 1000 1100 1200 3-10 1286 B.M. on boundary wall 14-96 21-18 Calculate the missing values and show the conventional arithmet- ical checks on your results. Reduced B.S. I.S. F.S. Rise Fall Level A.O.D. Distance Remarks 2-36 (a) 121-36 B.M. on Watson House 4-30 1-94 (b) 119-42 100 4-05 (c) 7-29 2-99 116-43 200 8-51 (d) 4-46 111-97 300 4-31 4-20 (e) 116-17 400 6-93 2-62 113-55 500 CO 7-16 0-23 113-32 600 (g) 4-80 7-79 0-63 112-69 715 Peg 36 3-22 1-58 114-27 800 (h) 5-48 2-26 0) 112-01 900 6-53 1-05 110-96 1000 (k) 3-96 2-57 113-53 1100 (m) (1) 3-75 5-86 (n) 1-90 111-63 1200 6-85 3-10 108-53 1286 B.M. on boundary wall 14-96 27-79 8-35 21-18 27-79 8-35 -12-83 -12-83 -12-83 254 SURVEYING PROBLEMS AND SOLUTIONS Notes: (a) 4-30 I.S. is deduced from fall 1-94. (b) Fall 2-99 is obtained from I.S.-F.S., 4-30-7-29. (c) 8-51 as (a). (d) Rise 4-20 as (b). (e) Fall 2-62 as (b). (f ) Reduced levels 113-32 - 113-55 gives fall 0-23, which gives staff reading 7-16. (g) 4-80 B.S. must occur on line of F.S. —deduced value from rise 1-58 with I.S. 3-22. (h) 5-48 as (f). (j) 1-05 as normal. (k) 3-96 as (f). (1) Fall 1-90 normal. (m) 3-75 B.S. must occur opposite 5-86 F.S. Value from 2b.S. 14-96. (n) 6-85 from fall 3-10. Checks as usual. Exercises 5(a) (Booking) 1. The undernoted staff readings were taken successively with a level along an underground roadway. Staff Readings Distances from A Remarks 5-77 B.S. to A 2-83 120 I.S. 0-30 240 F.S. 5-54 B.S. 1-41 360 I.S. 3-01 480 F.S. 2-23 B.S. 2-20 600 I.S. 1-62 720 F.S. 7-36 B.S. 5-52 840 I.S. 0-71 960 F.S. 4-99 B.S. 2-25 1080 F.S. to B Using the Height of Collimation method, calculate the reduced level of each staff station relative to the level of A, which is 6010*37 ft above an assumed datum of 10000ft below O.D. Thereafter check your results by application of the appropriate method used for verifying levelling calculations derived from heights of collimation. (M.Q.B./S Ans. Reduced level of B 6028*37) LEVELLING 255 2. A section levelling is made from the bottom of a staple pit A to the bottom of a staple pit B. Each group of the following staff read- ings relates to a setting of the levelling instrument and the appropriate distances from the staff points are given. From bottom of staple pit A. Staff Readings Distance (ft) 4-63 2-41 72 0-50 51 7-23 4-80 32 3-08 26 1-02 45 4-09 3-22 48 1-98 53 1-47 3-85 52 6-98 46 1-17 2-55 106 2-16 3-64 54 5-27 45 To bottom of staple pit B. The top of staple pit A is 8440ft above the assumed datum of 10000ft below O.D. and the shaft is 225ft deep. (a) Enter the staff readings and distances in level book form, com- plete the reduced levels and apply the usual checks. (b) Plot a section on a scale 100 ft to lin. for horizontals and 10 ft to lin. for verticals. (c) If the staple pit B is 187*4 ft deep what is the reduced level at the top of the shaft? (M.Q.B./UM Ans. 8404-85 ft) 3. Reduce the following notes of a levelling made along a railway affected by subsidence. Points A and B are outside the affected area, and the grade was originally constant between them. Find the original grade of the track, the amount of subsidence at each chain length and the maximum grade in any chain length. 256 SURVEYING PROBLEMS AND SOLUTIONS B.S. I.S. F.S. Distance (links) Remarks 10-15 8-84 100 A 7-58 6-65 7-69 200 300 4-21 2-72 5-50 400 500 8-21 3-76 2-38 5-55 600 700 Not on track 1-09 800 B (M.Q.B./M Ans. Grade 1*29 in 100 links; subsidence +0*02, -0*12, -0'48, -0-62, -0-42, -0-90 max. grade l'62ft in 100 links be- tween 500 - 600 ft) 4. The following staff readings were observed in the given order when levelling up a hillside from a temporary bench mark 135* 20 ft A.O.D. With the exception of the staff position immediately after the bench mark, each staff position was higher than the preceding one. Enter the readings in level book form by both the rise and fall and collimation systems. These may be combined, if desired, into a single form to save copying. 4*62, 8-95, 6-09, 3'19, 12*43, 9-01, 5*24, 1*33, 10*76, 6*60, 2-05, 13-57, 8-74, 3*26, 12'80, 6*33, 11-41, 4*37. (L.U.) N.B. There are alternative solutions 5. The undernoted readings, in feet, on a levelling staff were taken along a roadway AB with a dumpy level, the staff being held in the first case at a starting point A and then at 50ft intervals: 2*51, 3*49, [2-02], 6-02, 5-00. The level was then moved forward to another position and further readings taken. These were as follows, the last reading being at B: 7-73, 4-52, [6-77], 2'22, 6*65. The level of A is 137'2ft A.O.D. Set out the staff readings and complete the bookings. Calculate the gradient from A to B. (Figures in brackets denote inverted staff readings.) (R.I.C.S./M Ans. 1 in 284) 6. An extract from a level book is given below, in which various bookings are missing. Fill in the missing bookings and re-book and LEVELLING 257 complete the figures by the Height of Collimation method. B.S. I.S. F.S. Rise Fall Reduced Level 154-86 7-62 3-70 2-32 5-30 7-11 147-72 5-55 1-56 149-28 7-37 2-00 151-28 8-72 1-61 149-93 4-24 6-09 154-41 All the figures are assumed correct. (I.C.E.) 7. The following figures are the staff readings taken in order on a particular scheme, the backsights being shown in italics: 2'67, 7*12, 9-54, 8'63, 10-28, 12'31, 10*75, 6*23, 7*84, 9-22, 5-06, 4-18, 2' 11. The first reading was taken on a bench mark 129*80 O.D. Enter the readings in level book form, check the entries and find the reduced level of the last point. Comment on your completed reduction. (L.U./E) 5.4 Field Testing of the Level Methods available are (1) by reciprocal levelling, (2) the two-peg methods. 5.41 Reciprocal levelling method In Fig. 5.5, the instrument is first set at A of height a. The line of sight is assumed to be inclined at an angle of elevation + a giving an error e in the length AB. The reading on the staff at B is b. Difference in level A - B = a - (b-e) (5.4) Fig. 5.5 Difference in level 258 SURVEYING PROBLEMS AND SOLUTIONS Fig. 5.6 In Fig. 5.6, the instrument is set at B of height ft, and the reading on the staff at A is a r Difference in level A - B = (a,-e) - b, (5.5) Thus, from Eqs (5.4) and (5.5), Difference in level = a - b + e (5.4) = a, - fe, - e (5.5) Adding (5.4) and (5.5), = i[(a-b) + (a,-ft,)] (5.6) By subtracting Eqs. (5.4) and (5.5), the error in collimation e = i[(a,-5,)-(a-5)] (5.7) Example 5.6 A dumpy level is set up with the eyepiece vertically over a peg A. The height from the top of A to the centre of the eye- piece is measured and found to be 4*62 ft. A level staff is then held on a distant peg B and read. This reading is 2*12 ft. The level is then set over B. The height of the eyepiece above B is 4* 47 ft and a read- ing on A is 6* 59 ft. (1) What is the difference in level between A and B ? (2) Is the collimation of the telescope in adjustment? (3) If out of adjustment, can the collimation be corrected without moving the level from its position at 6? (I.C.E.) (1) From Eq. (5.6), Difference in level (A-B) = \ [(4- 62 -2-1 2) + (6* 59 -4* 47)] = \ [2-50 + 2-12] = +2-31 ft (2) From Eq. (5.7), Error in collimation e = | [2*12 - 2*50] = - 0-19 ft per length AB. (i.e. the line of sight is depressed.) LEVELLING 259 (3.) True staff reading at A (instrument at B) should be 6-59 -(-0-19) = 6-59 + 0*19 = 6-78 ft. The cross-hairs must be adjusted to provide this reading. 5.42 Two-peg method In the following field tests the true difference in level is ensured by making backsight and foresight of equal length. />-• a-b Fig. 5.7 Assuming the line of collimation is elevated by a, the displacement vertically = dtana Thus, if B.S. = F.S., dtana = e in each case. •*• True difference in level AB = (a-e)-(b-e) = a ~ b Method (a). Pegs are inserted at A and B so that the staff reading a = b when the instrument is midway between A and B. The instru- ment may now be moved to A or B. a,- 2»=^ Fig. 5.8 In Fig. 5.8, if the height of the instrument at B is fc, above peg B, the staff reading at peg A should be b, if there is no error, i.e. if a= 0. If the reading is a, and the distance AB = 2d, then the true read- 260 SURVEYING PROBLEMS AND SOLUTIONS ing at A should be a, - 2e :. e = |[a,-b,] (5.8) Method (b). The instrument is placed midway between staff positions A and B, Fig. 5.9. Readings taken give the true difference in level = a-b. a,-kt Fig. 5.9 The instrument is now placed at X so that XA = kXB where k is the multiplying factor depending on the ratio of AX/BX. If AX = kBX, then the error at A = ke True difference in level = (a y -ke) - (fe, -e) = a-b a,-6,-(&-l)e = a-b (a,-b,)-(a-fc) Error per length BX e = * ~ x (5.9) N.B. (1) If the instrument is placed nearer to A than B, k will be less than 1 and /c-1 will be negative (see Example 5.8). (2) If the instrument is placed at station B, then Eq. (5.9) is modified as follows: (a,-£) - b, = a-b .'. E = (a,-b,)-(a-b) (5.10) where E is the error in the length AB (see Example 5.9). Example 5.7 (a) When checking a dumpy level, the following readings were obtained in the two-peg test: Level set up midway between two staff stations A and B 400 ft apart; staff readings on A 5'75ft and on B 4*31 ft. Level set up 40 ft behind B and in line AB; staff reading on B 3-41 ft and on A 4*95 ft. LEVELLING 261 Complete the calculation and state the amount of instrumental error. (b) Describe the necessary adjustments to the following types of level, making use of the above readings in each case: (i) Dumpy level fitted with diaphragm screws and level tube screws . (ii) A level fitted with level screws and tilting screw. (a) By Eq. (5.9) e Check (4-95 - 3-41) - (5-75 - 4-31) 11 - 1 1-54 - 1-44 10 = 0*01 per 40 ft With instrument 40 ft beyond B, Staff reading at A should be 4*95 -(11 x 0-01) = 4-95- 0-11 = 4- 84 ft Staff reading at B should be 3'41 - 0*01 = 3'40ft Difference in level = 1*44 This agrees with the first readings 5*75 - 4'31 = 1'44 (b) (i) With a dumpy level the main spirit level should be first ad- justed. The collimation error is then adjusted by means of the dia- phragm screws until the staff reading at A from the second setting is 4*84 and this should check with the staff reading at B of 3*40. (ii) With a tilting level, the circular (pill-box) level should be first adjusted. The line of sight should be set by the tilting screw until the calculated readings above are obtained. The main bubble will now be off centre and must be central- ised by the level tube screws. Example 5.8 (a) Describe with the aid of a diagram the basic prin- ciples of a tilting level, and state the advantages and disadvantages of this type of level compared with the dumpy level. (b) The following readings were obtained with a tilting level to two staves A and B 200ft apart. 262 SURVEYING PROBLEMS AND SOLUTIONS Position of Instrument Midway between A and B 10 ft from A and 200 ft from B Reading at A (ft) 5-43 6-17 Reading at B(ft) 6-12 6-67 What is the error in the line of sight per 100 ft of distance and how would you adjust the instrument? (R.I.C.S. L/Inter.) Part (b) illustrates the testing of a level involving a negative angle of inclination of the line of collimation and a frac- tional k value. Using Eq. (5.9), 10 e - ("!-».>-("-»> where k- 1 (6-17-6-67)- (5-43 -6- 12) BX 200 1/20 - 1 20(- 0-50 +0-69) 19 = - 0-01 x 20 = - 0-2 ft per length BX Check i.e. e = - 0-2 per 200ft = - Q-l per 100ft At X, Reading on A should be 6*17 + 0*01 = 6*18 Reading on B should be 6*67 + 0-20 = 6*87 difference in level = - 0*69 Alternative solution from first principles (Fig.5.10) 6-67 -209 Fig. 5.10 True difference in level = (6'17-e) - (6'67-20e) = (5'43-6'12) i.e. 19e- 0-50 = -0*69 0-19 e = 19 = -0*01 per 10 ft = -0*1 ft per 100 ft LEVELLING 263 Example 5.9 The following readings taken on to two stations A and B were obtained during a field test of a dumpy level. Suggest what type of error exists in the level and give the magnitude of the error as a percentage. How would you correct it in the field? B.S. 6-21 4-99 By Eq. (5.10) F.S. 5-46 4-30 Remarks Staff at station A A-B 200ft apart Staff at station B Instrument midway Staff at station A AB 200 ft apart Staff at station B Instrument v. near to B (R.I.C.S./G) E = („,-&,) -(a-*) = (4-99-4-30) -(6-21-5-46) = 0-69-0-75 = - 0-06 ft per 200 ft = -0-03 ft per 100 ft. Check 4-99 + 0-06 = 4-30 + True difference in level = 5-05 4-30 0-75 ft 6-21 - 5-46 = 0-75 ft. Example 5.10 In levelling up a hillside to establish a T.B.M. (tem- porary bench mark) the average lengths of ten backsights and ten foresights were 80ft and 40ft respectively. As the reduced level of the T.B.M. of 82*50ft A.O.D. was in doubt, the level was set up midway between two pegs A and B 200 ft apart, the reading on A being 4*56 and that on B 5*24. When the instrument was moved 40ft beyond B on the line AB produced, the reading on A was 5*34 and on B 5*88. Calculate the true value of the reduced level. True difference in level A-B = 4-56-5*24 = -0*68 When set up 40 ft beyond B, (5'34-6e)-(5-88-e) = -0-68 5-34 - 5-88 - Se = - 0*68 Se = e = 0-14 0-028 ft/40 ft. 264 SURVEYING PROBLEMS AND SOLUTIONS Check True readings should have been: at A 5-34-0-168 = 5-172 at B 5-88 - 0-028 = 5-852 - 0-680 Error in levelling = 0*028 ft per 40 ft Difference in length between B.S. and F.S. per set = 80-40 = 40 ft per 10 sets = 400 ft .*• Error = 10 x 0*028 = - 0*28 ft .". True value of T.B.M. = 82'50 - 0'28 = 82-22 ft A.O.D. Exercises 5(b) (Adjustment) 8. Describe how you would adjust a level fitted with tribrach screws, a graduated tilting screw and bubble-tube screws, introducing into your answer the following readings which were taken in a 2 peg test: Staff stations at A and B 400ft apart. Level set up halfway between A and B : staff readings on A 4-21 ft, on B 2-82 ft. Level set up 40 ft behind B in line AB : staff readings on A 5-29 ft, on B 4*00 ft. Complete the calculation and show how the result would be used to adjust the level. (L.U. Ans. Error -0*01 per 40ft) 9. A modern dumpy level was set up at a position equidistant from two pegs A and B. The bubble was adjusted to its central position for each reading, as it did not remain quite central when the telescope was moved from A to B. The readings on A and B were 4*86 and 5*22 ft respectively. The instrument was then moved to D, so that the distance DB was about five times the distance DA, and the readings with the bubble central were 5*12 and 5*43 ft respectively. Was the instrument in adjustment? (I.C.E. Ans. Error = 0*0125 ft at A from D) 10. The table gives a summary of the readings taken when running a line of levels A, B, C, D. The level used was fitted with stadia hairs and had tacheome trie constants of 100 and 0. For all the readings the staff was held vertically. LEVELLING 265 Reduce the levels shown in the table Position of Backsight Foresight Staff Top Middle Bottom Top Middle Bottom A 6-22 4-37 2-52 B 4-70 2-94 1-18 11-06 9-38 7-70 C 7-63 5-27 2-91 9-32 7-43 5-54 D 8-17 6-04 3-91 It was suspected that the instrument was out of adjustment and to check this, the following staff readings were taken, using the centre hair of the level diaphragm; P and Q are 300 ft apart. Instrument Station Staff at P Staff at Q Near P Near Q 4-65 2-97 8-29 6-17 Find the true reduced level of D if the reduced level of A is 125-67 ft A.O.D. (Ans. 115-36 ft A.O.D.) 11. A level was set up on the line of two pegs A and B and readings were taken to a staff with the bubble central. If A and B were 150 metres apart, and the readings were 2*763 m and 1*792 m re- spectively, compute the collimation error. The reduced levels were known to be 27*002 m and 27*995 m respectively. The level was subsequently used, without adjustment, to level between two points X and Y situated 1 km apart. The average length of the backsights was 45 m and of the foresights 55 m. What is the error in the difference in level between X and Y? (N.R.C.T. Ans. 30*5" depressed; error + 0*0145m) 12. The following readings were taken during a 'two-peg' test on a level fitted with stadia, and reading on a vertical staff, the bubble being brought to the centre of its run before each reading. The points A, B, X and Y were in a straight line, X being mid- way between A and B and Y being on the side of B remote from A. (A) If the reduced level of A is 106*23, find the reduced level of B. (B) Explain what is wrong with the instrument and how you would correct it. (C) Find what the centre hair readings would have been if the instrument had not been out of adjustment. 266 SURVEYING PROBLEMS AND SOLUTIONS Instrument at Staff at Staff Readings X A 5-56 4-81 4-06 X B 8-19 7-44 6-69 Y A 5-32 3-72 2-12 Y B 6-31 6-21 6-11 (R.I.C.S./I VM. Ans. 103-60; 4-74, 7-37, 3*57, 6*20) 13. A level set up in a position 100 ft from peg A and 200 ft from peg B reads 6*28 on a staff held on A and 7*34 on a staff held on B, the bubble having been carefully brought to the centre of its run before each reading. It is known that the reduced levels of the tops of the pegs A and B are 287*32 and 286*35 ft O.D. respectively. Find (a) the collimation error and (b) the readings that would have been obtained had there been no collimation error. (L.U. Ans. (a) 0*09ft per 100ft, (b) 6*19; 7* 16ft) 14. P and Q are two points on opposite banks of a river about 100 yds wide. A level with an anallatic telescope with a constant 100 is set up at A on the line QP produced, then at B on the line PQ produced, and the following readings taken on to a graduated staff held vertically at P and Q. From To Staff Readings in ft Upper Stadia Collimation Lower Stadia P Q P Q 5-14 3-27 10-63 5-26 4-67 4-20 1-21 below ground 8-51 6-39 4-73 4-20 What is the true difference in level between P and Q and what is the collimation error of the level expressed in seconds of arc, there being 206 265 seconds in a radian? (I.C.E. Ans. 3*62ft; 104" above horizon) LEVELLING 267 5.5 Sensitivity of the Bubble Tube The sensitivity of the bubble tube depends on the radius of curva- ture (R) and is usually expressed as an angle (0) per unit division (d) of the bubble scale. 5.51 Field test Staff readings may be recorded as the position of the bubble is changed by a footscrew or tilting screw. Readings at the eye and ob- jective ends of the bubble may be recorded or alternatively the bubble may be set to the exact scale division. Reading (a) sac a— b Reading (M Fig. 5.11 In Fig. 5.11, tan(n0) = I but e is very small (usually 1 to 60 seconds) •'• "0rad = S I #rad = s nl d— = 206265 s (5.11) nl (5.12) (5.13) where s = difference in staff readings a and b n = number of divisions the bubble is displaced between readings / = distance from staff to instrument. If d = length of 1 division on the bubble tube, then d = R.d TBd R = * ndl s (5.14) (5.15) 268 SURVEYING PROBLEMS AND SOLUTIONS 5.52 - E correction If the bubble tube is graduated from the centre then an accurate reading is possible, particularly when seen through a prismatic reader, Fig. 5.12. Objective end Eye end 20 16 12 8 4 4 8 12 16 20 -+I-H — I — I — I — I — h — I — I — I — I — lr—f- I I I l i fl T Fig. 5.12 If the readings at the objective end are 0, and Z and those at the eye end E, and E 2 , then the movement of the bubble in n divi- sions will equal or (0 1 -E 1 ) + (0 2 -£ 2 ) 2 (0,+0 2 )-(E, + E 2 ) (5.16) The length of the bubble will be + E (5.17) — F The displacement of the bubble will be — - — (5.18) If > E the telescope is elevated, < E the telescope is depressed. 5.53 Bubble scale correction With a geodetic level, the bubble is generally very sensitive, say 1 division = 1 second. Instead of attempting to line up the prismatically viewed ends of LEVELLING 269 the bubble, their relative positions are read on the scale provided and observed in the eyepiece at the time of the staff reading. The correction to the middle levelling hair is thus required. ByEq. (5.13), R 206 265 s ''sec i nl Transposing gives e = "*£.,. (5.19) 206 265 where e - the error in the staff reading 6 = the sensitivity of the bubble tube in seconds n = the number of divisions displaced / = length of sight. Example 5.11 Find the radius of curvature of the bubble tube attached to a level and the angular value of each 2 mm division from the follow- ing readings taken to a staff 200 ft from the instrument. (2 mm = 0-006 56 ft). Staff Readings 3-510 3-742 Bubble Readings Eye End 18-3 6-4 Objective End 3-4 15-3 By Eq. (5.16), n = | [(18*3 - 3-4) + (15*3 - 6-4)] - |[14-9 + 8-9] = 11*9 divisions. By Eq. (5.13), 6 = 206265 s nl 206 265 x (3-742-3-510) 11-9 x 200 = 20 sec By Eq. (5.15), R = ndl s 0-006 56 x 200 x 11-9 0-232 = 67- 3 ft In the metric system the above readings would be given as: Staff readings l'070m M41m Distance between staff and level = 60- 96 m 270 SURVEYING PROBLEMS AND SOLUTIONS n*. L r, ,r,^ „ 206 265 x (1-141 -1-070) „. Then, by Eq. (5.13), - IT9TW96 = ^^ by Eq. (5.15), R . °-° 02 x^ 6 * "' 9 . 2H3« Example 5. 12 The following readings were taken through the eye- piece during precise levelling. What should be the true middle hair reading of the bubble value if 1 division is 1 second. The stadia con- stant of the level is x 100. Stadia Readings Top Middle Bottom 6-3716 5-507 4 4-643 1 By Eq. (5.18), n = ° " Bubble Scale Readings E O 10-6 8*48 Then by Eq. (5.19), e = 2 8-4 - 10-6 2 - 1-1 nW 206 265 - 1-1 x 100(6-3716- - 4-643 1) x 1" 206 265 110 x 1-7285 206265 = - 0*0009 .*. True middle reading should be 5' 507 4 + 0*0009 = 5-508 3 Exercises 5(c) (Sensitivity) 15. A level is set with the telescope perpendicular to two footscrews at a distance of 100 ft from a staff. The graduations on the bubble were found to be 0*1 in. apart and after moving the bubble through 3 divisions the staff readings differed by 0-029 ft. Find the sensitivity of the spirit bubble tube and its radius of curvature. (Ans. e ^ 20 seconds; R = 86*2 ft) 16. State what is meant by the term 'sensitivity' when applied to a spirit level, and discuss briefly the factors which influence the choice LEVELLING 271 of spirit level of sensitivity appropriate for the levelling instrument of specified precision. The spirit level attached to a levelling instrument contains a bubble which moves 1/10 in. per 20 seconds change in the inclination of the axis of the spirit level tube. Calculate the radius of curvature of the spirit level tube. (M.Q.B./S Ans. 85*94 ft) 5.54 Gradient screws (tilting mechanism) On some instruments the tilting screw is graduated as shown in Fig. 5.13. The vertical scale indicates the number of complete revolutions whilst the horizontal scale indi- cates the fraction of a revolution. The positive and negative tilt 10- of the telescope are usually shown in black and red respectively and these must be correlated with simi- lar colours on the horizontal scale. The gradient of the line of sight is given as 1 in x. where 1/x = nr (5.20) where n = numbers of revs. r = the ratio of 1 rev (fre- quently 1/1000). Fig. 5.13 mrrrrrm fio 40 -5 Vertical scale 30 20 v\ 50 60 70 80 90 Horizontal scale Using the gradient screw, it is also possible to obtain the approx- imate distance by taking staff readings. (5.21) If gradient = j- = nr then L = £ nr where s = staff intercept L = length of sight Example 5.13 Staff reading (a) = 6-32 00 = 6-84 Number of revs (n) = 6-35 Gradient ratio (r) = 1/1000 272 SURVEYING PROBLEMS AND SOLUTIONS (6-84 - 6-32) x 1 000 Then L = 6-35 520 6-35 = 81- 88 ft 5.6 The Effect of the Earth's Curvature and Atmospheric Refraction 5.61 The earth's curvature Over long distances the effect of the earth's curvature becomes significant. Let the error due to the earth's curvature E = AC. ■g Refracted line of sight In Fig. 5.14, ACAD = TA 2 .*. AC = TA 2 AD L 2 2R + AC i.e. E ~ L 2 2R Fig. 5.14 (intersecting chord and tangent) (where L = length of sight ~ TA) (as E is small compared with R) (5.22) Alternatively, by Pythagoras, AO 2 - 0T 2 +AT i.e. (E + Rf = R 2 +L 2 E 2 + 2RE + /T E R 2 +L' 2R + E ~ — as above. 2R LEVELLING 273 As R, the radius of the earth, is ^ 3 960 miles (^6 370 km). E ~ (5 280L) 2 2 x 5280 x 3960 5280L 2 7920 = 0-667 L 2 ft. (5.23) where L = length of sight in miles; or metric values give E = 0'0785 L 2 metres (where L = length in km) 5.62 Atmospheric refraction Due to variation in the density of the earth's atmosphere, affected by atmospheric pressure and temperature, a horizontal ray of light TA is refracted to give the bent line TB. If the coefficient of refraction m is defined as the multiplying factor applied to the angle TO A (subtended at the centre) to give the angle ATB, ^ Angle of refraction A TB = m TO A = 2m ATC As the angles are small AB : AC : : ATB : ATC Then AB = 2mA J C x AC ATC = 2m AC (5.24) The value of m varies with time, geographical position, atmos- pheric pressure and temperature. A mean value is frequently taken as 0-07. /. AB = 0-14 AC Error due to refraction ^ ±AC (5.25) ~ 0M567_L 2 = . 095L a (526 ) 5.63 The combined effect of curvature and refraction The net effect e = BC = AC - AB I 2 L z 2R 2R = Ll[l-2m] (5.27) ZK 274 SURVEYING PROBLEMS AND SOLUTIONS If m is taken as 0*07 then e = 0-667 L 2 (1 - 0- 14) = 0-667 L 2 x 0-86 = 0-574 L 2 ft (5.28) or metric value e m = 0*0673 L metres Alternatively, taking refraction as 1/7 of the curvature error, 6 7 = 0-572 L* - 0-57 V e = | x 0-667 L Example 5.14 Effect of curvature 1. What difference will exist between horizontal and level lines at the following distances? (a) 1 mile (d) 100 miles (b) 220 yards (e) 1 km (c) 5 miles (f) 160 km (a) E a = 0-667 L 2 ft (Lin miles) = 0-667 ft 8-004 inches (b) E b oc L 2 thus E b = 0-667 (|) 2 - 0*667 _ 0-010 4 ft 64 (c) E c = 0-667 x 5 2 25 x 0-667 66 ' 7 - 16-675 ft 4 (d) E d = 0-667 x 100 2 = 0-667 x 1000 6670 ft (e) Ee = 0-0785 x l 2 = 0-078 5 m (0 E f = 0-078 5 x 160 2 = 2 009-6 m In ordinary precise levelling it is essential that the lengths of the backsight and foresight be equal to eliminate instrumental error. This is also required to counteract the error due to curvature and refraction, as this error should be the same in both directions providing the clim- atic conditions remain constant. To minimise the effect of climatic change the length of sights should be kept below 150 ft. In precise surveys, where the length of- sight is greater than this value and climatic change is possible, e.g. crossing a river or ravine, 'reciprocal levelling' is employed. LEVELLING 275 Exercises 5(d) (Curvature and refraction) 17. Derive the expression for the combined curvature and refraction correction used in levelling practice. If the sensitivity of the bubble tube of a level is 20 seconds of arc per division, at what distance does the combined curvature and re- fraction correction become numerically equal to the error induced by dislevelment of one division of the level tube. (R.I.C.S. D/M Ans. 4 742*4 ft) 18. A geodetic levelling instrument which is known to be in adjust- ment is used to obtain the difference in level between two stations A and B which are 2430ft apart. The instrument is set 20ft from B on the line AB produced If A is 1*290 ft above B, what should be the reading on the staff at A if the reading on the staff at B is 4*055 ft. (M.Q.B./S Ans. 2*886ft) 5.64 Intervisibility The earth's curvature and the effect of atmospheric refraction affect the maximum length of sight, Fig. 5.15. ft, = 0*57 d z ft, = 0*57 (D - df Fig. 5.15 This will give the minimum height ft 2 at C which can be observed from A height ft, assuming the ray grazes the surface of the earth or sea. With intervening ground at B In Fig. 5.16, let the height of AA^ = ft, of BB 3 = h 2 = SB, + S,B 2 + B 2 B 3 = 5280dtana + ft, + 0*57 d' of CC 3 = h 3 = 5280 D tan £ + ft, + 0*57 D' If C is to be visible from A then a ^ /8. If a = /3, then tana = h 2 - /i, - 0*57d' 5280d ft, - 0-57P' 5280D (5.29) (5.30) (5.31) 276 SURVEYING PROBLEMS AND SOLUTIONS Fig. 5.16 Thus the minimum height d §-0-57.) K = jj(K -h,- 0-57 D 2 ) + h, + 0-57d ; - £*(D-d) Clendinning quotes the formula as Hz = 1) + ^ (D " d) " Kd( ° ~ d ) cosec2Z where K ~ 0*57 Z is the zenith angle of observation. Over large distances Z ^ 90° .'. cosec 2 Z ^ 1. Example 5.15 If h x = 2 300 ft (at A), d = 46 miles h 2 = 1050 ft (at B), D = 84 miles h 3 = 1800 ft (at C). Can C be seen from /I ? By Eq. (5.32), „ 2 = 1800x46 + (g4 84 = 985-7 + 44-1 = 1029-8 ft 46) ( 2300 84 0-57 x 46 (5.32) (5.33) LEVELLING 277 2300' 54-8' 1800' Fig. 5.17 The station C cannot be seen from A as h z is > 1029*8. If the line of sight is not to be nearer than 10 ft to the surface at B, then it would be necessary to erect a tower at C of such a height that the line of sight would be 10 ft above B, 84 i.e. so that its height h = (1 050 + 10 - 1 029*8) = 30*2 x 1*826 ~ 54* 8 ft 46 Exercises 5(e) (Intervisibility) 19. Two ships A and B are 20 miles (32* 18 km) apart. If the observer at A is 20 ft (6* 096 m) above sea level, what should be the height of the mast of B above the sea for it to be seen at A? (Ans. 113* Oft (34* 5 m)) 20. As part of a minor triangulation a station A was selected at 708*63 ft A.O.D. Resection has been difficult in the area and as an additional check it is required to observe a triangulation station C 35 miles away (reduced level 325*75 ft A.O.D.). If there is an inter- mediate hill at B, 15 miles from A (spot height shown on map near B 370 ft A.O.D.), will it be possible to observe station C from A as- suming that the ray should be 10 ft above B? (N.R.C.T. Ans. Instrument + target should be ~ 20 ft) 21. (a) Discuss the effects of curvature and refraction on long sights as met with in triangulation, deriving a compounded equation for their correction. (b) A colliery headgear at A, ground level 452*48 ft A.O.D., is 145ft to the observing platform. It is required to observe a triangulation station C, reduced level 412*68 ft A.O.D., which is 15 miles from A, but it is thought that in- tervening ground at B approximately 500ft A.O.D. and 5 miles from A 278 SURVEYING PROBLEMS AND SOLUTIONS will prevent the line of sight. Assuming that the ray should not be nearer than 10 ft to the ground at any point, will the observation be possible? If not, what height should the target be at C? (R.I.C.S./M Ans. >5ft) 22. Describe the effect of earth's curvature and refraction on long sights. Show how these effects can be cancelled by taking reciprocal observations. Two beacons A and B are 60 miles apart and are respectively 120 ft and 1200 ft above mean sea level. At C, which is in the line AB and Is 15 miles from B,. the ground level is 548 ft above mean sea level. Find by how much, if at all, B should be raised so that the line of sight from A to B should pass 10 ft above the ground at C. The mean radius of the earth may be taken as 3960 miles. (L.U. Ans. +17 ft) 5.65 Trigonometrical levelling For plane surveying purposes where the length of sight is limited to say 10 miles the foregoing principles can be applied, Fig. 5.18. — "I?* 5280d tanoC 1 0-57of z 1 • Fig. 5.18 The difference in elevation = h. h, = 5 280dtana+0'57d (5.34) where d = distance in miles a = angle of elevation. If the distance D is given in feet, then A, - A, - Dtana + 0-57 (^ = Dtana + 2'04 x 10~ 8 D 2 (5.35) N.B. It is considered advisable in trigonometrical levelling, and in normal geometrical levelling over long distances, to observe in both directions, simultaneously where possible, in order to eliminate the effects of curvature and refraction, as well as instrumental errors. This is known as reciprocal levelling. LEVELLING 279 Example 5.16 The reduced level of the observation station A is 350*36 ft A.O.D. From A, instrument height 4*31 ft, the angle of ele- vation is 5° 30' to station B, target height 6*44 ft. If the computed distance AB is 35 680* lft what is the reduced level of B? Reduced level of B = Reduced level of A + difference in eleva- tion + instrument height - target height By Eq. (5.35), Difference in elevation = 35 680*1 tan 5 ° 30' + 2*04 x 10~ 8 x 35 680' l 2 - 3 435-60 + 25-97 = 3 461- 57 ft Reduced level of B = 3 461*57 + 4*31 - 6- 44 = 3 459- 44 ft A.O.D. Based on metric values the problem becomes: The reduced level of the observation station A is 106'790m A.O.D. From A, instrument height 1-314 m, the angle of elevation is 5° 30' to station B, target height 1*963 m. If the computed distance AB is 10-875 3 km, what is the reduced level of B? Difference in elevation = 10 875*3 tan 5 ° 30' + 0*067 3 (10*875 3 2 ) = 1047-172 + 7-960 = 1055- 132 m Reduced level of B = 1055-132 + 1-314 - 1*963 = 1054* 483m (3 459* 59 ft) A.O.D. 5.7 Reciprocal Levelling Corrections for curvature and refraction are only approximations as they depend on the observer's position, the shape of the geoid and atmospheric conditions. To eliminate the need for corrections a system of Reciprocal Levelling is adopted for long sights. In Fig. 5.19, Difference in level = d = fiX, = a^ + c + e-r-by from A = (a, - b x ) + (c - r) + e Also from B d = AX 2 = - (b 2 + c + e - r ~ a 2 ) = (« 2 ~b 2 ) - (c-r) - e By adding, 2d = (a, - ft,) + (a 2 - b 2 ) d = |[(a,-ft.) + (a 2 -b 2 )] (5.36) 280 SURVEYING PROBLEMS AND SOLUTIONS Uneo^g^f-^^^li^ Horizontal line Line due to refractTol? ~-^-^e, H ■fll b 2 - L|n£_o'_sight_due to col/fmati, Horizontal line i : ;^Tw retraction ?T^ Line due ^ ^^- Fig. 5.19 Subtracting, 2(c - r + e) = [(a 2 - b 2 ) - (a, - b,)] Total error (c - r) + e = ±[(a 2 - b 2 ) - (a, - &,)] (5.37 By calculating the error due to refraction and curvature, Eq. (5.28), for (c-r) the collimation error e may be derived. (See §5.4.) Example 5.17 (a) Obtain from first principles an expression giving the combined correction for earth's curvature and atmospheric refrac- tion in levelling, assuming that the earth is a sphere of 7 920 miles diameter. (b) Reciprocal levelling between two points Y and Z 2 400ft apart on opposite sides of a river gave the following results: LEVELLING Instrument at Height of instrument Staff at Staff reading Y 4-80 Z 5-54 Z 4-71 y 3-25 281 Determine the difference in level between Y and Z and the amount of any collimation error in the instrument. (I.C.E.) (a) By Eq. (5.28), c ~ 0'57d 2 (b) By Eq. (5.36), Difference in level = |[(a, - ft,) + (a 2 - b 2 )] = i[(4-80 - 5-54) + (3-25 - 4*71)] = \ [-0-74 - 1-46] = - MO ft i.e. Z is MOft below Y By Eq. (5.37), Total error (c - r) + e = i [(a 2 ~ b z ) ~ (a, - &,)] = |[- 1*46 + 0-74] = - 0-36 ft By Eq. (5.28), (c- i.e -r) r^ e = 0-57d 2 . 57 /2400\ 2 V5 280/ 0-118 ft - 0-478 ft per 2 400 ft - 0-02 ft per 100 ft (collimation depressed) Cfcecfc Difference in i level = 4-80 - - 5-54 - 0-48 + 0-12 = - MOft also 3-25 - 4-71 + 0-48 - 0-12 = - MOft 5.71 The use of two instruments To improve the observations by removing the likelihood of clim- atic change two instruments should be used, as in the following example. 282 SURVEYING PROBLEMS AND SOLUTIONS Example 5.18 Instrument Staff at A Mean Staff at B Mean Apparent Difference in level Remarks I L 6-934 9-424 Inst. I on same M 6-784 6-784 8-072 8-073 -1-289 side as A U 6-634 6-722 n L 6-335 6-426 Inst. II on same M 4-985 4-984 6-276 6-276 -1-292 side as B U 3-633 6-126 I L 6-452 6-514 Inst. I on same M 5-098 5-099 6-364 6-364 -1-265 side as B U 3-747 6-214 II L 6-782 9-249 Inst. II on same M 6-632 6-632 7-893 6-895 -1-263 side as A U 1 6-482 6-543 4 ) -5-109 True difference in level - 1*277 Thus B is 1-277 ft below A. Exercises 5(f) (Reciprocal levelling) 23. The results of reciprocal levelling between stations A and B 1 500ft apart on opposite sides of a wide river were as follows: Level at Height of Eyepiece (ft) 4-59 4-37 Staff Readings 8-26 on B 1-72 on A Find (a) the true difference in level between the stations (b) the error due to imperfect adjustment of the instrument assum- ing the mean radius of the earth 3 956 miles. (L.U./E Ans. (a) -3-16 ft; (b) + 0' 031 ft /100ft) 24. In levelling across a wide river the following readings were taken: Instrument at Staff Reading at A 5- 98 ft (l-823m) 8- 20 ft (2- 499 m) Staff Reading at B 8- 14 ft (2-481 m) 10- 44 ft (3- 182 m) If the reduced level at A is 102*63 ft (31*282 m) above datum what is the reduced level of B? (Ans. 100*43 ft (30*612 m)) LEVELLING 283 5.8 Levelling for Construction 5.81 Grading of constructions The gradient of the proposed construction will be expressed as 1 in x, i.e. 1 vertical to x horizontal. The reduced formation level is then computed from the reduced level of a point on the formation, e.g. the starting point, and the pro- posed gradient. By comparing the existing reduced levels with the proposed re- duced levels the amount of cut and fill is obtained. If formation > existing, fill is required. If formation < existing, cut is required. Example 5.19 The following notes of a sectional levelling were taken along a line of a proposed road on the surface. B.S. I.S. F.S. Height of Collimation Reduced Level Horizontal Distance Remarks 10-24 4-63 1-47 104-52 100 B.M. Station 1 Station 2 8-52 5-23 0-41 200 300 Station 3 Station 4 12-64 3-37 5-87 400 500 Station 5 Station 6 Calculate the reduced level of each station and apply the conven- tional arithmetical checks. Thereafter calculate the depth of cutting and filling necessary at each station to form an even gradient rising at 1 in 20 and starting at a level of 105 ft above datum at station 1. (M.Q.B./M) B.S. I.S. F.S. Height of Collimation Reduced Level Formation Level Cut Fill Station 10-24 114-76 104-52 4-63 110-13 105-00 5-13 1 1-47 113-29 110-00 3-29 2 8-52 0-41 122-87 114-35 115-00 0-65 3 5-23 117-64 120-00 2-36 4 12-64 3-37 132-14 119-50 125-00 5-50 5 5-87 126-27 701-18 130-00 3-73 6 31-40 11-33 9-65 9-65 21-75 284 SURVEYING PROBLEMS AND SOLUTIONS NB. Vertical scale exaggerated (x10 horizontal scale) - CM CO * in Chainage 0) * 1 8 8 Existing level o 6 o o 6 CM o to CM o 8 Formation level o> CM <*> Cut IS 6 <0 CO CM o in in CO CO Fill Fig. 5.20 Check 701-18 + 11*33 + 9*65 = 722-16 (114*76 x 3) + (122-87 x 2) + (132-14 x 2) = 722-16 5.82 The use of sight rails and boning (or travelling) rods Sight rails and boning rods are used for excavation purposes as- sociated with the grading of drains and sewers. The sight rails are established at fixed points along the excava- tion line, at a height above the formation level equal to the length of the boning rod. The formation level compared with the surface level gives the depth of excavation, Fig. 5.21. Boning rod Sight rail Fig. 5.21 LEVELLING 285 When the boning rod is in line with the sight rails the excavation s at the correct depth, Fig. 5.22. Sight rail Boning rod ^Jimmmmsmm^mmm^mm Length of boning rod Formation^ oradie_t_ Sight rail Length of boning rod Fig. 5.22 Example 5.20 In preparing the fixing of sight rails, the following consecutive staff readings were taken from one setting of the level: Benchmark (165*65ft A.O.D.) 2*73 Ground level at A 5*92 Invert of sewer at A 10* 63 Ground level at B 4*27 Ground level at C 3*54 If the sewer is to rise at 1 in 300 and the distance AB 105 ft and BC 153 ft, what will be the height of the sight rails for use with 10 ft boning rods? What is the reduced level of the invert at A, B and C? 2-73 165-65 A.O.D. ! 5-92 (10-63 invert) 1Q5' 3-54 153' 168-38'A.O.D. Invert of •ew-rJJP. *mVmVMW;WM\WW/AWMWM«> 300 Fig. 5.23 In Fig. 5.23, Height of collimation = 165*65 + 2* 73 Invert of sewer at A = 168*38 - 10*63 Sight rail at A = 157*75 + 10*00 Ground level at A = 168*38 - 5*92 Height of sight rail above ground at A 168*38 ft A.O.D. 157*75 ft 167*75 ft 162*46 ft 5*29 ft 286 SURVEYING PROBLEMS AND SOLUTIONS Gradient of sewer 1 in 300 Invert of sewer at B = Invert at A + rise due to gradient = 157-75 + 105/300 = 158- 10 ft Sight rail at B = 168* 10 ft Ground level at B = 168'38 - 4' 27 = 164- 11 ft Height of sight rail above ground at B = 3' 99 ft Invert of sewer at C = Invert at B + rise due to gradient = 158-10+153/300 = 158-61 ft Sight rail at C = 158-61 + 10-00 = 168*61 ft Ground level at C = 168-38 - 3' 54 = 164-84 ft 3- 77 ft 5. S3 The setting of slope stakes A slope stake is set in the ground at the intersection of the ground and the formation slope of the cutting or embankment. The position of the slope stake relative to the centre line of the formation may be obtained: (a) by scaling from the development plan, or (b) by calculation involving the cross-slope of the ground and the formation slope, using the rate of approach method suggested in §8.3. Formation level Fig. 5.24 By the rate of approach method, in Fig. 5.24, w h < ■ A ° + w ft, = K 2K (5.38) (5.39) LEVELLING 287 By Eq. (8.14), «*, - ■*, 1 1 hi ~ k Similarly, d 2 - K l l (5.40) (5.41) M K Point A Point B 5- 63 ft 6- 13 ft Example 5.21 To determine the position of slope stakes, staff read- ings were taken at ground level as follows: Centre line of proposed road (Reduced level 103' 72 ft A.O.D.) 50 ft from centre line and at right angles to it If the reduced level of the formation at the centre line is to be 123-96 A.O.D., the formation width 20ft, and the batter is to be 1 in 2, what will be the staff reading, from the same instrument height at the slope stake and how far will the peg be from the centre line point A? 123-96 Fig. 5.25 Gradient of AB = (6*13- 5*63) in 50ft i.e. 0-5 in 50ft 1 in 100 In Fig. 5.25, AA, = 123*96 - 103*72 20 By Eq. (5.38), XX, = fc = 20*24 + = 20*24 ft = 20*34 ft 2x 100 By Eq. (5.40), the horizontal distance <2,i.e. XP, h 20*34 1 _ 1 M K 2 100 100 x 20*34 50 - 1 = 41*51 ft 288 SURVEYING PROBLEMS AND SOLUTIONS The distance from the centre line point A = 51* 51 ft the inclined length XP = 41-51 x VUOO 2 + 1) 100 = 41-72 ft Level of A = 103-72 Difference in level AP - 41 ' 5 } + 10 100 = 0-52 ft Level of P = 103-20 Height of collimation = 103*72 + 5*63 = 109-35ft Staff reading at P = 6- 15 ft Exercises 5(g) (Construction levelling) 25. Sight rails are to be fixed at A and B 350 ft apart for the setting out of a sewer at an inclination of 1 in 200 rising towards B. If the levels of the surface are A 106*23 and B 104*46 and the invert level at A is 100*74, at what height above ground should the sight rails be set for use with boning rods 10 ft long? (Ans. 4*51 at A; 8'03 at B) 26. A sewer is to be laid at a uniform gradient of 1 in 200 between two points X and Y, 800 ft apart. The reduced level of the invert at the outfall X is 494*82. In order to fix sight rails at X and Y, readings are taken with a level in the following order: Reading Staff Station B.S. 2-66 T.B.M. (near X) Reduced level 504-64 I.S. a Top of sight rail at X I.S. 3-52 Peg at X F.S. 1-80 T.P. between X and Y B.S. 7-04 T.P. between X and Y I.S. b Top of sight rail at Y F.S. 6-15 Peg at Y (i) Draw up a level book and find the reduced levels of the pegs, (ii) If a boning rod of length 9'- 6" is to be used, find the readings a and b. (iii) Find the height of the sight rails above the pegs at X and Y. (L.U. Ans. (ii)2*98, 4*22; (iii) 0- 54, 1*93) 27. The levelling shown on the field sheet given below was under- taken during the laying out of a sewer line. Determine the height of LEVELLING 289 the ground at each observed point along the sewer line and calculate the depth of the trench at points X and Y if the sewer is to have a gradient of 1 in 200 downwards from A to B and is to be 4*20 ft below the surface at A. B.S. I.S. F.S. Distance (ft) Remarks 11-21 B.M. 321-53 4-56 3-78 5-82 100 11-65 3-66 200 Point X 2-40 3-57 300 7-82 5-91 10-81 400 500 6-56 600 Point Y 6-32 8-65 3-81 700 Point B B.M. 329-15 (R.I.C.S./M/L Ans. 6*10, 9'03) Exercises 5(h) (General) 28. The following staff readings in fact were taken successively with a level, the instrument having been moved forward after the second, fourth and eighth reading. 1*54, 7*24, 4'03, M5, 8'62, 8'52, 6*41, 1-13, 7-31, 2-75 and 5'41. The last reading was taken with the staff on a bench mark having an elevation of 103-74 ft. Enter the readings in level book form, complete the reduced levels and apply the usual checks. 29. The following readings were taken using a dumpy level on a slightly undulating underground roadway. B.S. I.S. F.S. Reduced Level Distance Remarks 5-32 6-43 5-23 + 8 752-20 100 200 Point A + 8 752-20 3-06 4-30 2-23 4-12 300 400 500 3-02 4-01 5-12 1-09 600 700 800 6-67 900 Point B Work out the reduced levels relative to the assumed datum of mean 290 SURVEYING PROBLEMS AND SOLUTIONS sea level +10 000 ft (as used by the National Coal Board to avoid negative reduced levels). State the amount of excavation necessary at point B to form an even gradient dipping 1 in 300 from A to B, the reduced level of A to remain at 8 752*20 ft. (Ans. 2- 52 ft) 30. Reduce the page of a level book and plot the result to a scale of 1 = 100' horizontal and 1" = 10' vertical. B.S. I.S. F.S. Rise Fall Reduced Level Distance Remarks 9-92 8-22 5-98 25-23 100 B.M. Start of Section 7-35 5-59 3-15 200 300 8-13 6-05 5-63 5-00 2-12 400 460 500 560 5-63 4-19 5-91 3-65 600 630 700 4-71 5-35 4-01 8-04 800 830 900 6-24 5-82 4-36 2-73 3-72 1000 1100 1200 End of Section B.M. (R.I.C.S./Q) 31. A levelling party ran a line of levels from point A at elevation 135*43 to point B for which the reduced level was found to be 87*15. A series of flying levels (as below) was taken back to the starting point A. B.S. F.S. Remarks 9-67 B 11-54 1-38 8-22 4-81 7-94 3-35 10-56 2-07 9-92 5-33 8-88 1-04 0-42 A LEVELLING 291 Find the misclosure on the starting point. (L.U./E Ans. 0'05ft) 32. (a) Explain the difference between 'rise and fall' and 'height of collimation' method of reducing levels, stating the advantages and dis- advantages of each. (b) The following is an extract from a level book. Reduce the levels by whichever method you think appropriate, making all the necessary checks and insert the staff readings in the correct blank spaces for setting in the levels pegs A, B and C so that they have the reduced levels given in the book. B.S. I.S. F.S. Reduced Level Distance Remarks 3-24 58-63 B.M. 1 5-03 9-65 C.P. 6-42 Beginning of Se 9-69 50 - 49-69 50 Peg A 4-19 10-87 100 C.P. - 48-55 100 Peg B 5-54 150 - 47-41 150 Peg C 4-30 200 End of Section 11-73 2-32 C.P. 8-61 51-35 B.M. 2 (R.I.C.S./M/L Ans. Staff Readings, A 7'56, B 2*02, C 3-16; Error in levelling, 0*02) 33. The level book refers to a grid of levels taken at 100 ft intervals on 4 parallel lines 100ft apart. (A) Reduce and check the level book. (B) Draw a grid to a scale of 50 ft to lin. and plot the contours for a 2 ft vertical interval. B.S. I.S. F.S. Reduced Level Distance on Line Remarks 1-23 2-75 3-51 4-26 89-14 100 200 300 Line A T.B.M. 12-35 300 Line B 9-06 200 6-78 100 4-18 4-15 5-51 6-97 100 Line C 292 SURVEYING PROBLEMS AND SOLUTIONS B.S. LS. F.S. Reduced Level Distance on Line Remarks 7-88 200 10-45 300 8-93 300 Line D 7-18 200 5-34 100 4-59 8-37 2-62 4-14 C.P. T.B.M. (R.I.C.S./L/M) 34. Discuss the various ways in which 'errors' can occur in levelling and measures that can be adopted to keep each source of error to a minimum. In levelling from a bench mark 347*79 ft above O.D. and closing on to another 330*61 ft above O.D., staff readings were taken in the fol- lowing order: 3*72, 8*21; 4*91, 8*33, 7*28; 0*89, 4*27; 2*28, 3*91, 3*72, 9*23. The position of the instrument was moved immediately after taking the 2nd, 5th, and 7th readings indicated by semi-colons in the above series of readings. Show how these readings would be booked and the levels reduced using either the 'collimation' or the 'rise and fall' method. Carry out the usual arithmetical checks and quote the closing error. Explain briefly why it is particularly important not to make a mis- take in reading an intermediate sight. /T _ „ N (l.U.L.) 35. The record of a levelling made some years ago has become of cur- rent importance. Some of the data are undecipherable but sufficient re- main to enable all the missing values to be calculated. Reproduce the following levelling notes and calculate and insert the missing values. B.S. I.S. F.S. Rise Fall Reduced Level Remarks 2-36 1-94 121-36 B.M. at No. 1 Shaft 4-05 4-31 6-93 3-22 6-53 7-29 7-79 1-58 4-46 0-63 113-32 112-01 B.M. on School LEVELLING 293 B.S. I.S. F.S. Rise Fall Reduced Level Remarks 5-86 3-10 21-18 113-53 B.M. on Church 14-96 36. The following is an extract from a level book B.S. I.S. F.S. Reduced Level Remarks 4-20 2-70 Point A 2-64 3-42 9-51 11-74 11-40 119-30 C.P. B.M. 119-30 2-56 3-10 6-91 13-75 C.P. B 3-61 5-60 11-23 C.P. C 12-98 3-61 C.P. C 13-62 3-31 C.P. B 12-03 2-51 4-83 C.P. B.M. Point A (a) Reduce the above levels. (b) If you consider a mistake has been made suggest how it occurred. (c) Give reasons for choice of 'Rise and Fall' or 'Height of Col- limation' for reducing the levels. The B.S. and F.S. lengths were approximately equal. (L.U. Ans. probably 11*98 instead of 12*98) 37. The following are the levels along a line ABC. Distance Reduced Level Remarks 0-00 At A 10 1-21 20 2-46 30 3-39 40 4-54 At B 50 6-03 60 7-65 70 9-03 80 10-32 At C 294 SURVEYING PROBLEMS AND SOLUTIONS Plot the reduced levels to a scale of 10 ft to lin. for the horizon- tal scale and 1ft to lin. for the vertical scale. A roadway is to be constructed from A to C at a uniform gradient. From the section state the height of filling required at each plotting point. (R.I.C.S./M) 38. In order to check the underground levellings of a colliery it was decided to remeasure the depth of the shaft and connect the levelling to a recently established Ordnance Survey Bench Mark A, 272*45 ft above O.D. The following levels were taken with a dumpy level starting at A to the mouth of the shaft at D. B.S. F.S. Reduced Level Remarks 2-17 272-45 ft B.M. at A 3-36 11-32 Mark B 5-79 7-93 Mark C 0-00 Mark D on rails The vertical depth of the shaft was then measured from D to E at the pit bottom and found to be 1 745 ft 8^ in. A backsight underground to E was found to be 3*98 and a fore- sight to the colliery Bench Mark F on a wall near the pit bottom was 2-73. Tabulate the above readings and find the value of the underground B.M. at F expressing this as a depth below Ordnance Datum. (Ans. 1479-94 ft) 39. Levels were taken at 100 ft intervals down a road with a fairly uniform gradient and the following staff readings booked: B.S. I.S. F.S. Distance 7-00 9-50 100 T.B.M. 98-50 11-75 8-55 10-81 6-00 200 300 400 13-05 7-90 10-20 5-60 500 600 700 12-70 8-48 10-98 13-20 6-00 800 900 1000 1 100 LEVELLING 295 Errors were made in booking, correct these and reduce the levels. (L.U. Ans. B.S. and F.S. transposed) 40. Spot levels are given below at 200 ft intervals on a grid ABCD. Draw the plan to a scale of 200 ft to 1 in. and show on it where you would place the level in order to take readings. Draw up a level book by the 'height of collimation' method show- ing your readings. Take the level at A as a T.B.M. A 100-70 102-00 103-50 105*20 106-80 108-20 109-50 B 101-30 103-40 104-10 106'30 108*20 109-30 110-70 105-00 106-20 107-30 109-10 110-40 111-50 112*30 D 108-00 107-10 108-60 110-40 111-30 112*20 113-80 C (L.U.) 41. In levelling up a hillside, the sight lengths were observed with stadia lines, the average length of the ten backsights and foresights being 70 ft and 35 ft respectively. Since the observed difference of the reduced level of 78* 40 ft was disputed, the level was set up midway between two pegs A and B 300 ft apart, and the reading on A was 4*60 and on B 5*11; and when set up in line AB, 30 ft behind B, the reading on A was 5*17 and on B 5-64. Calculate the true difference of reduced level. (L.U. Ans. 78-35 ft; 0*013 ft per 100ft) 42. A, B, C, D, E and F are the sites of manholes, 300 ft apart on a straight sewer. The natural ground can be considered as a plane sur- face rising uniformly from A to F at a gradient of 1 vertically in 500 horizontally, the ground level at A being 103*00. The level of the sewer invert is to be 95*00 at A, the invert then rising uniformly at 1 in 200 to F. Site rails are to be set up at A, B, C, D, E and F so that a 10ft boning rod or traveller can be used. The backsights and foresights were made approximately equal and a peg at ground level at A was used as datum. Draw up a level book showing the readings. (L.U.) 43. The following staff readings were obtained when running a line of levels between two bench marks A and B: 3*56 (A) 6*68, 7-32, 9*89 change point, 2*01, 6*57, 7*66, C.P. 5*32, 4*21, 1-78, C.P. 4*68, 5*89, 2*99 (S) Enter and reduce the readings in an accepted form of field book. The reduced levels of the bench marks at A and B were known to be 143-21 ft and 136*72 ft respectively. It is found after the readings have been taken with the staff 296 SURVEYING PROBLEMS AND SOLUTIONS supposedly vertical, as indicated by a level on the staff, that the level is 5 ° in error in the plane of the staff and instrument. Is the collimation error of the instrument elevated or depressed and what is its value in seconds if the backsights and foresights averaged 100ft and 200 ft respectively. (L.U. True difference in level 6*72; collimation elevated 119sec) 44. Undernoted are levels taken on the floor of an undulating under- ground roadway AB, 10 ft in width and 6 ft in height, which is to be regraded and heightened. Distance (ft) B.S. I.S. F.S. Rise Fall Reduced Level Remarks 6-95 30-00 Floor level at A 50 3-40 100 0-65 150 2-50 0-00 200 3-50 250 6-00 300 6-50 350 7-75 5-00 400 4-50 450 1-25 500 1-00 2-20 6-65 6-55 Floor level at B Floor level at A Plot the section of the roadway to a scale of lin. = 50 ft for hori- zontals and lin. = 10ft for verticals. Thereafter calculate and mark on the section the amount of ripping and filling at the respective 50 ft intervals to give a uniform gradient from A to B and a minimum height of 8 ft. Calculate the volume, in cubic yards, of the material to be ripped from the roof in giving effect to the above improvements. (M.Q.B./M Ans. 320yd 3 ) 45. The centre-line of a section of a proposed road in cutting is indi- cated by pegs at equal intervals and the corresponding longitudinal section gives the existing ground level and the proposed formation level at each peg, but no cross-sections have been taken, or sidelong slopes observed. Given the proposed formation width (d) and the batter of the sides (S horizontal to 1 vertical) how would you set out the batter pegs marking the tops of the slopes at each centre line peg, without taking and plotting the usual cross-sections? An alternative method would be acceptable. (I.C.E.) LEVELLING 297 46. (a) Determine from first principles the approximate distance at which correction for curvature and refraction in levelling amounts to 0*01 ft, assuming that the effect of refraction is one seventh that of the earth's curvature and that the earth is a sphere of 7 920 miles di- ameter. (b) Two survey stations A and B on opposite sides of a river are 2510 ft apart, and reciprocal levels have been taken between them with the following results: Instrument at Height of instrument Staff at Staff reading (ft) A B 4-83 4-91 B A 6-02 3-98 Compute the ratio of refraction correction to curvature correction, and the difference in level between A and B. (LOE. Ans. (a) ~ 700ft; (b) A is 1-06 ft above B. Ratio ~ 0*14 to 1) Bibliography HOLLAND, J.L., WARDELL, K. and WEBSTER, A.G., Surveying, Coal Mining Series (Virtue). SANDOVER, J. A. Plane Surveying (Edward Arnold). TRUTMAN, O., Levelling (Wild (Heerbrugg)). PARRY, R. and JENKINS, W.R., Land Surveying (Estates Gazette). BANNISTER, A. and RAYMOND, S., Surveying (Pitman). CURTEN, w. and LANE, R.F., Concise Practical Surveying (English Universities Press). O TRAVERSE SURVEYS The purpose of traverse surveys is to control subsequent detail, i.e. the fixing of specific points to which detail can be related. The accuracy of the control survey must be superior to that of the subsi- diary survey. A traverse consists of a series of related points or stations, which when connected by angular and linear values form a framework. 6.1 Types of Traverse 6.11 Open Traverse ABCDE, Fig.6.1. The start and finish are not fixed points. c Fig. 6. 1 Open traverse A check on the angles may be made by (a) taking meridian observa- tions at the start and finish or (b) taking observations to a common fixed point X. 6.12 Closed (a) On to fixed points. If the start and finish are fixed points, e.g. A and E, then the length and bearing between them is known. From the traverse the distance may be computed. (b) Closed polygon, Fig. 6.2. Checks (i) The sum of the deflection angles should equal 360°, i.e. 2a = a, + a 2 + a 3 ...a n = 360° = 4 x 90° or (ii) The sum of the internal angles should equal (2« - 4) x 90° where n = no. of angles or sides, i.e. S/3 = jQ, + /S 2 + p 3 .../3 n = (2n -4)90° (6.1) or (iii) The sum of the external angles should equal 298 TRAVERSE SURVEYS 299 (2n + 4) 90° Le - 20 = 0, + 2 + 3 ...0 n = (2n + 4)90° (6.2) «4 Fig. 6.2 Closed traverse N.B. a, + /3, = a 2 + 2 = a n + n = 2 x 90° 2(a + 0) = (4 x 90) + (2n - 4)90 = (2 x 90) x n. The sum will seldom add up exactly to the theoretical value and the 'closing error' must be distributed before plotting or computing. 6.2 Methods of Traversing The method is dependent upon the accuracy required and the equip- ment available. The following are alternative methods. (1) Compass traversing using one of the following: (a) a prismatic compass (b) a miners' dial (c) a tubular or trough compass fitted to a theodolite (d) a special compass theodolite. (2) Continuous azimuth (fixed needle traversing) using either (a) a miners' dial or (b) a theodolite. (3) Direction method using any angular measuring equipment. (4) Separate angular measurement (double foresight method) using any angular measuring equipment. 300 SURVEYING PROBLEMS AND SOLUTIONS 6.21 Compass traversing (loose needle traversing), Fig. 6.3 Application. Reconnaissance or exploratory surveys. Advantages. (1) Rapid surveys. (2) Each line is independent — errors tend to compensate. (3) The bearing of a line can be observed at any point along the line. (4) Only every second station needs to be occupied (this is not recommended because of the possibility of local attraction) Disadvantages. (1) Lack of accuracy. (2) Local attraction. Accuracy of survey. Due to magnetic variations, instrument and obser- vation errors, the maximum accuracy is probably limited to ± lOmin, i.e. linear equivalent 1 in 300. Detection of effects of local attraction. Forward and back bearings should differ by 180° assuming no instrumental or personal errors exist. Elimination of the effect of local attraction. The effect of local attrac- tion is that all bearings from a given station will be in error by a con- stant value, the angle between adjacent bearings being correct. Where forward and back bearings of a line agree this indicates that the terminal stations are both free of local attraction. Thus, starting from bearings which are unaffected, a comparison of forward and back bearings will show the correction factors to be applied. X source of attraction Fig. 6.3 Compass traversing In Fig. 6.3 the bearings at A and B are correct. The back bearing of CB will be in error by a compared with the forward bearing BC. TRAVERSE SURVEYS 301 The forward bearing CD can thus be corrected by a. Comparison of the corrected forward bearing CD with the observed back bearing DC will show the error /3 by which the forward bearing DA must be cor- rected. This should finally check with back bearing AD. Example 6.1 Line Forward Bearing Back Beari AB 120° 10' 300° 10' BC 124° 08' 306° 15' CD 137° 10' 310° 08 ' DE 159° 08' 349° 08' EF 138° 15' 313° 10' Station Back Bearing Forward Bearing Correction Corrected Forward Bearing ± 180° A 300° 10 ' 120° 10* 124° 08' - B 124° 08' 304° 08' C 306° 15' 137° 10' -2° 07' 135° 03' 315° 03' D 310° 08' 159° 08' + 4° 55' 164° 03' 344° 03 ' E 349° 08' 313° 10' 138° 15' -5° 05' 133° 10 ' 313° 10' F Thus stations A, B, and F are free from local attraction. N.B. (1) Line AB. Forward and Back bearings agree; therefore sta- tions A and B are free from attraction (2) Corrected forward bearing at B + 180° compared with back bearing at C shows an error of 2° 07', i.e. 304° 08' - 306° 15' = -2° 07'. (3) Corrected forward bearing CD 137° 10' - 2° 07' = 135° 03'. (4) Comparison of corrected forward bearing EF + 180° agrees with back bearing FE. Therefore station F is also free from local attraction 6.22 Continuous azimuth method (Fig. 6.4) This method was ideally suited to the old type of miners' dial with open-vane sights which could be used in either direction. The instrument is orientated at each station by observing the back- sight, with the reader clamped, from the reverse end of the 'dial' sights. The recorded value of each foresight is thus the bearing of each line relative to the original orientation. For mining purposes this was the magnetic meridian and hence the method was known as 'the fixed needle method*. 302 SURVEYING PROBLEMS AND SOLUTIONS The method may be modified for use with a theodolite by changing face between backsight and foresight observations. Fig. 6.4 Continuous azimuth method of traversing 6.23 Direction method The continuous azimuth method of traversing is restrictive for use with the theodolite where the accuracy must be improved, as the obser- vations cannot be repeated. To overcome this difficulty, and still retain the benefit of carrying the bearing, the Direction method may be employed. This requires only approximate orientation, corrections being made either on a 'Direction' bearing sheet if one arc on each face is taken, or, alternatively, on the field booking sheet. N.B. No angles are extracted, the theodolite showing approximate bearings of the traverse lines as the work proceeds. Direction bearing sheet Set! Obs. at to A E B B A C C B D D C E E D A Mean observed directions 08-00 283 09.05 103 00.60 345 37*05 165 36-40 039 40-05 219 55-50 101 31-35 281 31-20 180 15-10 Corr- ection + 8-45 + 8-45 + 9-10 + 9-10 -6-35 -6-35 -6-20 -6-20 Back bearing 08-00 103 09-05 165 45-50 219 49-15 281 25-00 Initial bearing j Error Forward bearing 283 09-05 345 45-50 039 49-15 101 25-00 180 08-90 180 08-00 0-90 Final correction -0-18 -0-36 -0-54 -0-72 -0-90 Final bearing 08-00 283 08-87 345 45-14 039 48-61 101 24-28 180 08-00 TRAVERSE SURVEYS 303 N.B. (1) Here the instrument has been approximately orientated at each station, i.e. the reciprocal of the previous forward bearing is set as a back bearing. Any variation from the previous mean forward bear- ing thus requires an orientation correction. (2) In the closed traverse the closing error is seen immediately by comparing the first back bearing with the final forward bearing. (3) As a simple adjustment the closing error is distributed equally amongst the lines. Method of booking by the direction method Station set at A Back bearing AE C °08' Arc Obs. to FL. F.R. Mean Correction Back bearing Forward bearing o ' o ' ' ' ' ' 1 E 08-0 180 08-0 08-00 B 283 09-0 103 09-1 283 09-05 2 E 090 12-2 270 12-2 090 12-20 -90 04-20 08-00 B 013 13-2 193 13'4 013 13-30 -90 04-20 283 09-10 3 E 27-6 27-4 27-50 19-50 08-00 B 28-5 28*5 28-50 - 19-50 283 09-00 Station set at B Mean Forward Bearing 283 09-05 Back bearing BA 103° 09-05* Arc Obs. to F.L. F.R. Mean Correction Back bearing Forward bearing ' ' ' ' ' ' 1 A 103 00-6 283 00-5 103 00-55 + 08-50 103 09-05 C 345 37-1 165 37*2 345 37-15 + 08-50 345 45-65 2 A 199 04-9 019 05-0 199 04-95 -95 55-90 103 09-05 C 081 41-3 261 41-5 081 41-40 -95 55-90 345 45-50 3 A 15-6 15-7 15-65 - 06-60 103 09-05 C 52-0 52-2 52-10 06-60 345 45-50 Mean Forward Bearing 345 45-55 N.B. (1) No angles are extracted. (2) The back bearing is approximately set on the instrument for the backsight, e.g. AE is set exactly here 0°08-0' .*• AB is the correctly oriented bearing 283° 09-0' . (3) On arc 2 the instrument zero is changed. After taking out the mean of the faces, a correction is applied to give the back bearing, i.e. 090° 12-20' - 0° 08-00' Correction 090° 04-20' This correction is now applied to give the second forward bearing. 304 SURVEYING PROBLEMS AND SOLUTIONS (4) On the 3rd and subsequent arcs, if required, only the min- utes are booked, a new zero being obtained each time. (5) The mean forward bearing is now extracted and carried for- ward to the next station. 6.24 Separate angular measurement Angles are measured by finding the difference between adjacent recorded pointings. After extracting the mean values these are converted into bearings for co-ordinate purposes. In the case of a closed polygon, the angles may be summated to conform with the geometrical properties, Eq. (6.1) or (6.2). Exercises 6(a) 1. A colliery plan has been laid down on the national grid of the Ordnance Survey and the co-ordinates of the two stations A and B have been converted into feet and reduced to A as a local origin. Departure (ft) Latitude (ft) Station 4 Station B East 109-2 South 991-7. Calculate the Grid bearing of the line AB. The mean magnetic bearing of the line AB is S 3° 54 'W and the mean magnetic bearing of an underground line CD is N 17°55'W. State the Grid bearing of the line CD. (M.Q.B./M Ans. 331° 54') 2. The following angles were measured in a clockwise direction, from the National Grid North lines on a colliery plan: (a) 156° 15' (b) 181° 30' (c) 354° 00' (d) 17° 45' At the present time in this locality, the magnetic north is found to be 10° 30 'W of the Grid North. Express the above directions as quadrant bearings to be set off using the magnetic needle. (M.Q.B./UM Ans. (a)S13°15'E; (b)S12°00'W; (c)N4°30'E; (d)N28°15'E) 3. The following underground survey was made with a miners' dial in the presence of iron rails. Assuming that station A was free from local attraction calculate the correct magnetic bearing of each line. Station BS FS A 352° 00' B 358° 30' 12° 20' C 14° 35' 282° 15' TRAVERSE SURVEYS 305 Station BS FS D 280° 00' 164° 24' E 168° 42' 200° 22' (Ans. 352°00'; 05°50'; 273°30'; 157°54'; 189°34') (N.B. A miners' dial has vane sights, i.e. B.S. = F.S., not reciprocal bearings). 4. The geographical azimuth of a church spire is observed from a tri- angulation station as 346° 20'. At a certain time of the day a magnetic bearing was taken of this same line as 003° 23'. On the following day at the same time an underground survey line was magnetically observed as 195° 20' with the same instrument. Calculate (a) the magnetic declination, (b) the true bearing of the underground line. (Ans. 17°03'W; 178° 17') 5. Describe and sketch a prismatic compass. What precautions are taken when using the compass for field observations? The following readings were obtained in a short traverse ABC A. Adjust the readings and calculate the co-ordinates of B and C if the co-ordinates of A are 250 ft E, 75 ft N. Line Compass bearing Length (ft) AC 00° 00' 195-5 AB 44° 59' 169-5 BA 225° 01' 169-5 BC 302° 10' 141-7 CB 122° 10' 141-7 CA 180° 00' 195-5 (R.I.C.S. Ans. B 370E, 195N C 250E, 270N) 6. The following notes were obtained during a compass survey made to determine the approximate area covered by an old dirt-tip. Correct the compass readings for local attraction. Plot the survey to a scale of 1 in 2400 and adjust graphically by Bowditch's rule. Thereafter find the area enclosed by equalising to a triangle. Line Forward bearing Back bearing Length (ft) AB N57°10'E S 58°20'W 750 BC N81°40'E S 78°00'W 828 CD S 15°30'E N 15°30'W 764 DE S 10°20'W N 12°00'E 405 EF S 78°50'W N76°00'E 540 FG N69°30'E S 68°30'W 950 GA N 22° 10'W S 19°30'E 383 (N.R.C.T. Ans. AB 54°40'; BC 78°00'; CD 164° 30'; D£190°20'; EF257°10'; FG 291° 40'; GA 338° 00'; area 32-64 acres) 306 SURVEYING PROBLEMS AND SOLUTIONS 7. A and B are two reference stations in an underground roadway, and it is required to extend the survey through a drift from station B to a third station C. The observations at B were as follows: Horizontal angle ABC 271° 05' 20". Vertical angle to staff at C + 10° 15' 00". Staff reading at C 1-50 ft. Instrument height at B 5ft 7 in. Measured distance BC 284-86 ft. The bearing of AB was 349° 56' 10" and the co-ordinates of B E 4689-22 ft, N 5873-50 ft. Calculate the true slope of BC to the nearest 10 seconds, the horizontal length of BC, its bearing, and the co-ordinates of C. (N.R.C.T. Ans. 11°06'20"; 279-55 ft; 081°01'30"; E 4965-35, N 5917-11) 6.31 6.3 Office Tests for Locating Mistakes in Traversing A mistake in the linear value of one line If the figure is closed the co-ordinates can be computed and the closing error found. Linear mistake Fig. 6.5 Location of a linear mistake Let the computed co-ordinates give values for ABC X D^A V Fig. 6.5. The length and bearing of AA y suggests that the mistake lies in this direction, and if it is parallel to any given line of the traverse this is where the mistake has been made. The amount AA X is therefore the linear mistake, and a correction to the line BC gives the new station values of C, D and thus closes on A. If the closing error is parallel to a number of lines then a repetition of their measurements is suggested. TRAVERSE SURVEYS 6.32 A mistake in the angular value at one station Let the traverse be plotted as ABCD^A V Fig. 6.6 307 Station where mistake occurred Fig. 6.6 Location of an angular mistake The closing error AA y is not parallel to any line but the perpendi- cular bisector of /L4, when produced passes through station C. Here an angular mistake exists. Proof. AA) represents a chord of a circle of radius AC, the perpendi- cular bisector of the chord passing through the centre of the circle of centre C. The line CD, must be turned through the angle a = ACA V 6.33 When the traverse is closed on to fixed points and a mistake in the bearing is known to exist The survey should be plotted or computed from each end in turn, i.e. ABCDE - E,D X C,BA, Fig. 6.7. The station which is common to both systems will suggest where the mistake has been made. Fig. 6.7 308 SURVEYING PROBLEMS AND SOLUTIONS If there are two or more mistakes, either in length or bearing, then it is impossible to locate their positions but they may be localised. 6.4 Omitted Measurements in Closed Traverses Where it is impossible to measure all the values (either linear, angular or a combination of both) in a closed traverse, the missing quantities can be calculated provided they do not exceed two. As the traverse is closed the algebraic sum of the partial co-ordin- ates must each sum to zero, i.e. /,sin0, + / 2 sin0 2 + / 3 sin0 3 + ... / n sin0 n = /,cos0, + f 2 cos0 2 + J 3 cos0 3 + ... I n cos0 n = where the lengths of the lines are /,/ 2 / 3 etc -» ana< the bearings 0,0 2 3 etc. As only 2 independent equations are involved only 2 unknowns are possible. Failure to close the traverse in any way transfers all the traverse errors to the unknown quantities. Therefore use of the process is to be deprecated unless there is no other solution. Six cases may arise: (1) Bearing of one line. (2) Length of one line. (3) Length and bearing of one line. (4) Bearing of two lines. (5) Length of two lines. (6) Bearing of one line and length of another line. Cases 1, 2 and 3 are merely part of the calculation of a join between two co-ordinates. 6.41 Where the bearing of one line is missing / n sin0 n = P (1) where P = the sum of the other partial departures / n cos d n = Q (2) where Q = the sum of the other partial latitudes Dividing (1) by (2), P AE tan0 n = — i.e. jrzi = the difference in the total co- ^ ordinates of the stations form- ing the ends of the missing line (6.3) TRAVERSE SURVEYS 309 6.42 Where the length of one line is missing / n sin0 n = P (1) / n cos0 n = Q (2) By squaring each and adding / 2 sin 2 n = P 2 / n 2 cos 2 n = Q 2 '. / n 2 (sin 2 n + cos 2 6 n ) = P 2 + 2 i- e - k = V(P 2 +Q*) = y/(AE 2 + AN 2 ) (6.4) AE = iinT < 65 > AN cos6 n 6.43 Where the length and bearing of a line are missing The two previous cases provide the required values. 6.44 Where the bearings of two lines are missing (1) // the bearings are equal l p sind p + l q sind Q = P l p cos d p + l Q cos6 q = Q *% = e q = Then l p smd + l g sin$ = P or «* + lg)sin0 = P sin 6 = P l v + la cos 6 = Q l p + U tan# = P Q (2) // the bearings are adjacent /,sin#, + / 2 sin0 2 = P /,cos<9, + l 2 cosd 2 = Q (6.6) (6.7) 310 SURVEYING PROBLEMS AND SOLUTIONS In Fig.6.8, /,, l 2 , Z 3 , l A , <9 3 ,and 6 4 are known. AC = l s can be found with the bearing AC. In triangle ABC, tan ° _ fa-bKs-.c) 2 "V S (s - a) where sinS a + b + c 2 b sin a Fig. 6.8 From the value of the angles the required bearings can be found. Bearing AB = bearing AC - angle a Bearing BC = bearing BA - angle B (3) // the bearings are not adjacent Fig. 6.9 Assume 6 ' and 6 are missing. Graphical solution (Fig. 6.9) Plot the part of the survey in which the lengths and bearings are known, giving the relative positions of A and D. At A draw circle of radius AB - /,; this gives the locus of station B. At D draw circle of radius DC = l 3 ; this gives the locus of station C. From A plot length and bearing l 2 6 2 to give line AH. At H draw arcs HC X and HC 2 , radius /,, cutting the locus of station C at C, and C 2 . At C, and C 2 draw arcs of radius BC - l 2 , cutting the other locus at B, and B z - TRAVERSE SURVEYS 311 Mathematical solution Using the graphical solution: Find the length and bearing AD. Solve triangle AHD to give HD. Solve triangle HC^D to give <f> and /3 and thence obtain the bearings of tfC, = bearing AB X , HC 2 = bearing AB Z , C,D and C Z D. N.B. There are two possible solutions in all cases (1), (2) and (3), and some knowledge of the shape or direction of the lines is required to give the required values. Alternative solution Let /,sin0, + / 3 sin0 3 = P (1) /,cos0, + / 3 cos 3 = Q (2) i.e. ^sintf, = P - / 3 sin0 3 (3) /,cos0, = Q - / 3 cos0 3 (4) Squaring (3) and (4) and adding, If = P 2 + Q z + \\ - 2/ 3 (P sin 3 + Q cos 3 ) P Vtf^+Q 2 ) sin $3 + V(P 2 + Q 2 ) C0S ° 3 = P 2 + Q z +H-l? 2l 3 y/(P 2 +Q 2 ) Referring to Fig. 6.10 sin a sin0_ + cos a cos0 a = P 2 + Q 2 + l\ - I 2 2l 3 yJ(P 2 +&) i.e. cos(0 3 - a) = k 6 3 - a = cos"'/: a = tan rP Q = k (6.8) from (3), a = cos -1 fc + tan -1 — Q . P - / 3 sin0_ SU10, = 2 § (6.9) (6.10) Example 6.2 The following data relate to a closed traverse ABCD in which the bearings of the lines AB and CD are missing. 312 SURVEYING PROBLEMS AND SOLUTIONS Length Bearing AB 200 - BC 350 102° 36 ' CD 150 - DA 400 270° 00' Calculate the missing data. Method (1) Fig. 6. 11 s. Fig. 6.11 In triangle ADE, AE = BC = 350 AD = 400 By co-ordinates relative to A, st.(D) 400 E ON st.(E) 350 sinl02°36' = +350 sin77°24' = +341-57 350 cos 102° 36' = - 350 cos 77° 24' = - 76-34 tan bearing ED _ ™ - 34 ^ 57 _ «*43 + 76-34 76-34 bearing ED = N37°26'E = 037° 26' length ED = 76-34 sec 37° 26' = 96-14 58-43 cosec37°26' = 96-13 {check) In triangle EDC, where s ED + DC + EC I [ (126-93) (73 -07) ) A/l(223-07)(23-07)j <f>/2 = 53° 19' cf> = 106° 38' check 126-93 73-07 23-07 s 223-07 TRAVERSE SURVEYS 313 DC sin<£ sin/8 = EC 150 sin 106° 38' 200 /3 = 45° 56' Bearing AB 2 = 037° 26' + 45° 56' = 083° 22' or AB, = 037° 26' - 45° 56' = 351° 30' Bearing DC, = 217° 26' + 106° 38' = 324° 04' or DC 2 = 217° 26' - 106°38' = 110° 48' .". Bearing CD = 144° 04' or 290° 48' Method (2) 200 sin 0, + 350 sin 102° 36' + 150 sin 2 + 400 sin 270° = (1) 200cos<9, + 350cosl02°36' + 15Ocos0 2 + 400cos270° = (2) i.e. 200 sin 0, + 150 sin 2 = -341*57 + 400 = 58*43 (3) 200 cos 0, + 150 cos 2 = 76*34 + = 76*34 (4) .*. 200 sin0, = 58*43 - 150 sin0 2 (5) 200 cos 0, = 76*34 - 150 cos 2 (6) Squaring and adding, 200 2 = S8-43 2 + 76*34 2 + 150 2 - 2 x 150 (58*43 sin 2 + 76*34 cos 2 ) • cq^q • a koa a 58*43 2 + 76*34 2 + 150 2 -200 2 • • 58*43 sin O + 76*34 cos 0, = cos (0 2 - a) 300 5S-43 2 + 76*34 2 + 150 2 - 200 2 300 V(58*43 2 + 76*34 2 ) 3414*07 + 5827*79 + 22500 - 40 000 300 V(3414*07 + 5827*79) 2 - a = -73° 22' = 106° 38' or 253° 22' . 4 t 58*43 but tana = 76*34 a - 37° 26' ••• 0, = 144° 04' or 290° 48' 314 from (5) SURVEYING PROBLEMS AND SOLUTIONS 58*43- 150 sin 144° 04' sin 6. = 200 0, = 351° 30' or sin#, = *, 58-43 -150 sin 290° 48' 200 83° 21' 6.45 Where two lengths are missing Let /,sin0, + / 2 sin0 2 = P /,cos#, + / 2 cos0 2 = Q (a) The simultaneous equations may be solved to give values for / and l 2 regardless of their position. (b) If they are adjacent lines the solution of a triangle ADE will give the required values (Fig. 6.12), as length AD together with angles a and /3 are obtainable* Fig. 6.12 ~"X> e * (c) If 0, = 6 2 =6 (i.e. the lines are parallel) (/,+ / 2 )sin0 = P (/,+ / 2 )cos0 = Q Squaring and adding, </ f +/ a ) a = P 2 + Q Z Therefore this is not determinate. (d) If /, = l 2 = I and 6, = d 2 = 6 2/sin0 = P P / = 2 sin<9 (6.11) ♦The lines can be adjusted so that the missing values are adjacent. Solution (b) can then be applied. TRAVERSE SURVEYS 315 6.46 Where the length of one line and the bearing of another line are missing I Let /,sin0, + / 2 sin0 2 = P /,cos0, + / 2 cos0 2 = Q where 0, and l 2 are missing. Then, as before, /, sin0, = P - l 2 sin0 2 (1) Z,cos0, = Q - / 2 cos0 2 (2) Square and add, I 2 = P 2 + Q 2 + l\- 2/ 2 (Psin0 2 + Qcos0 2 ) this resolves into a quadratic equation in l 2 . I 2 - 2/ 2 (P sin0 2 + Q cos0 2 ) + P 2 + Q 2 - if = (6.12) Then from the value of Z 2 , 0, may be obtained from equation (1). Example 6.3 Using the data of a closed traverse given below, calcul- ate the lengths of the lines BC and CD. Line Length (ft) W.C.B. Reduced bearing Latitude Departure N 14°31'E +333-0 + 86-2 N 40° 18' W N 12°45'W N 05°16'E +298*8 + 27-6 S 11°48'E -1916-4 +400-4 (I.C.E.) Assuming the co-ordinates of A to be + 1000 E, + 1000 N, from the given co-ordinates: AB 344 014° 31' BC 319° 42' CD 347° 15' DE 300 005° 16' EA 1958 168° 12' *A + 1000-0 N* + 1000-0 &£ AE - 400-4 ^ae + 1916-4 *E + 599-6 N* 2916-4 &E ED - 27-6 ^ED - 298-8 *D + 572-0 K D 2617-6 Ea + 1000-0 K A + 1000-0 ^AB + 86-2 ^AB + 333-0 E* + 1086-2 N* + 1333-0 AE™ - 514-2 AN« n + 1284*6. -"BD Jlii ^"BD 316 SURVEYING PROBLEMS AND SOLUTIONS Bearing BD = tan"'- 514*2/+ 1284'6 = N21°49'W = 338° 11' BC = 319° 42' Angle CBD = 18° 29' DB = 158° 11' DC = 167° 15' Angle BDC = 9° 04' (fi + D) = 27° 33' Length BD = 1284'6/cos21°49' = 1383-7 In triangle BCD, DB sin B dn 9 e (B) (D) DC = sin(B + D) BC = 1383-7 sin 18° 29' sin 27° 33' = 948-4 ft DB sin D 1383-7 sin 9° 04' sin(B + D) sin 27° 33' 471 -4 ft 18° 29' Exercises 6(b) (Omitted values) 8. A clockwise traverse ABCDEA was surveyed with the following results: AB 331-4 ft EAB 128° 10' 20" BCD 84° 18' 10" BC 460-1 ft CD 325-7 ft ABC 102° 04' 30" CDE 121° 30' 30" TRAVERSE SURVEYS 317 The angle DEA and the sides DE and EA could not be measured direct. Assuming no error in the survey, find the missing lengths and their bearings if AB is due north. (L.U. Ans. EA = 223-lft, DE = 293-7ft, 308° 10' 20", 232° 06' 50" ) 9. An open traverse was run from A to E in order to obtain the length and bearing of the line AE which could not be measured direct, with the following results: Line AB BC CD DE Length 1025 1087 925 1250 W.C.B. 261° 41' 9° 06' 282° 22' 71° 30' Find by calculation the required information. (L.U. Ans. 1901; 342° 51') 10. The following measurements were obtained when surveying a closed traverse ABCDEA: Line EA AB BC Length (ft) 793-7 1512-1 863-7 DEA EAB ABC BCD Included angles 93° 14' 122° 36' 131° 42' 95° 43' It is not possible to occupy D, but it could be observed from both C and E. Calculate the angle CDE, and the lengths CD and DE, taking DE as the datum, and assuming all observations to be correct. (L.U. Ans. 96° 45'; 1848-0 ft, 1501-6 ft) 11. In a traverse ABCDEFG, the line BA is taken as the reference meridian. The latitudes and departures of the sides AB, BC, CD, DE and EF are: Line AB BC CD DE EF Latitude - 1190*0 - 565*3 + 590'5 + 606*9 + 1097*2 Departure + 736*4 + 796*8 - 468*0 + 370*4 If the bearing of FG is N 75°47'W and its length is 896'Oft, find the length and bearing of GA. (L.U. Ans. 947-6 ft; S 36°45'W) 6.5 The Adjustment of Closed Traverses (a) Traverses connecting two known points. (b) Traverses which return to their starting point. 6.51 Where the start and finish of a traverse are fixed The length and bearing of the line joining these points are known and must be in agreement with the length and bearing of the closing line of the traverse. 318 SURVEYING PROBLEMS AND SOLUTIONS Where the traverse is not orientated to the fixed line an angle of swing (a) has to be applied. Where there is discrepancy between the closing lengths, a scale factor k must be applied to all the traverse lengths. k = length between the fixed points closing length of traverse Fig. 6.14 In Fig. 6.14, the traverse is turned through angle a so that traverse ABCD becomes AB^D and AD } is orientated on to line XY. The XY scale factor k ^~rf\ must be applied to the traverse lines. Traverses are often orientated originally on their first line. Co- ordinates are then computed, and from these the length and bearing of the closing line (AD). The latter is then compared with the length and bearing XY. Co-ordinates of the traverse can now be adjusted by either (1) recomputing the traverse by adjusting the bearings by the angle a and the length by multiplying by k, or (2) transposing the co-ordinates by changing the grid (Eqs. 3.33/3.34) and also applying the scale factor k, or (3) applying one of the following adjustment methods, e.g. Bowditch. N.B. The factor k can be a compounded value involving: (a) traverse error, (b) local scale factor — (ground distance to national grid), (c) change of units, e.g. feet to metres. Example 6.4. A traverse XaY is made between two survey stations X (E 1000 N 1000) and Y (E 1424*5 N754'9). TRAVERSE SURVEYS 319 Based upon an assumed meridian, the following partial co-ordinates are computed: AE AN Xa 69-5 - 393*9 aY 199-3 - 17-4 Adjust the traverse so that the co-ordinates conform to the fixed stations X and Y. Ans. E N X 1000-0 1000-0 Y 1424-5 754-9 AE + 424*5 AN - 245*1 Bearing of control line XY = tan"' 424*5/- 245*1 = S 60° OO'E i.e. 120° 00' Length of control line XY = 424*5/sin60° = 490*2. From the partial co-ordinates, AE XY = 69*5 + 199*3 - + 268*8 AN^ y = -393*9- 17-4 = -411*3 Bearing of traverse line XY = tan -1 + 268*8/- 411*3 = S33°10 , E i.e. 146° 50' Length of traverse line XY = 411*3/cos33° 10' = 491*4. Angle of swing a = traverse bearing XY - fixed bearing XY = 146° 50' - 120° 00' = + 26° 50' Scale factor k = fixed length/traverse length = 490*2/491*4 = 0*99756 Using Eqs.(3.33) and (3.34), AE' = + mAE - nAN AN' = mAN + «AE m = kcosa = 0*99756 cos 26° 50' = 0*89014 n = ksina = 0*99756 sin 26° 50' = 0*45030 320 SURVEYING PROBLEMS AND SOLUTIONS Line Xa aY Ae An + 69-5 -393-9 + 199-3 - 17-4 mAE nAN + 61-86 -177-37 + 177-40- 7-84 Total co-ordinates X a Y Ae' + 239-23 + 185-24 + 424-47 E 1000-0 1239-2 1424-5 + mAN nAE -350-63 +31-30 - 15-48 +89-74 N 1000-0 680-7 754-9 An' -319-33 + 74-26 -245-07 Example 6.5 If in the previous example the co-ordinates of Y are E 1266-9 N 589-1, then the bearing of XY = tan -1 + 266*9/ - 410-9 = S33°00'E i.e. 147° 00' length XY = 410-9/cos33° - 490-0 angle of swing a = 146° 50' - 147° 00' = -0°10' a fad = -0-002 91 Scale factor k = 490*0/491-4 = 0-99716 Using Eqs. (3.35) and (3.36), AE' = fc[AE - ANa] AN' = fc[AN + AEa] Line Xa aY Ae An + 69-5 -393-9 + 199-3 - 17-4 ANa AEa + 1-15 -0-20 + 0-05 -0-58 Ae - ANa An + AEa + 68-35 -394-10 + 199-25 - 17-98 Ae' An' + 68-16 -392-98 + 198-68 - 17-93 2 +266-84 -410-91 Example 6.6 An underground traverse between two wires in shafts A and D based on an assumed meridian gives the following partial co- ordinates: AE(ft) AN (ft) AB 263-516 BC + 523-684 + 21*743 CD + 36-862 +421*827 If the grid co-ordinates of the wires are: TRAVERSE SURVEYS 321 A D E (metres) 552361-63 552 532-50 N (metres) 441 372-48 441428-18 Transform the underground partials into grid co-ordinates. Arts. Bearing of traverse line AD = tan" 1 + 560*546/ + 180-054 - N72°ll'33"E Length of traverse line AD = 560*546 /sin 72° 11 '33" = 588*755 ft Bearing of grid line AD = tan" 1 170-87/55*70 = N71°56'45"E Length of grid line AD = 170*87/sin71° 56'45" = 179*713 m Angle of swing a = 72°11'33" - 71°56'45" = +0°14'48" a rad = 0*004 31. Scale factor k = 179*713/588*755 = 0*30523 Using Eqs.(3.35) and (3.36), AE' = fc[AE - ANa] AN' = fc[AN + AEa] Line AE AN ANa AEa AE - ANa AN + AEa AE' AN' AB o-o - 263-516 - 1-136 + 1-136 -263-516 + 0-35 - 80-43 BC + 523-684 + 21-743 + 0-094 +2-257 + 523-590 + 24-000 + 159-82 + 7-33 CD + 36-862 + 421-827 + 1-818 + 0-159 + 35-044 +421-986 + 10-70 + 128-80 X +170-87 + 55-70 E N A 552 361 -63 m 441 372-48 m B 552 361-98 m 441 292-05 m C 552 521 -80 m 441 299-38 m D 552 532-50 m 441 428-18 m Example 6.7 Fig. 6.15 shows a short 'dial' traverse connecting two theodolite lines in a mine survey. The co-ordinates of T.M.64 are E 45603-1 ft N 35 709-9 ft and of T.M.86 E 46 163*6 ft N 35 411*8 ft. Cal- culate the co-ordinates of the traverse and adjust to close on T.M.86. (N.R.C.T.) 241° 26' 216° 57' 251° 06' Fig. 6.15 T.M.87 322 SURVEYING PROBLEMS AND SOLUTIONS This is a subsidiary survey carried out with a 1 minute instrument (miners' dial) and the method of adjustment should be as simple as possible. Bearing T.M. 63- T.M. 64 048° 19' + angle 241° 26' 289° 45' i Adjusted Bearings -180° Bearing T.M. 64 -(1) 109° 45' + 01' 109° 46' S70°14' E + angle 216° 57' 326° 42' - 180° Bearing 1-2 146° 42' + 02' 146° 44' S33°16 E + angle 110° 41' 257° 23' - 180° Bearing 2 -T.M. 86 077° 23' + 03' 077° 26' N77°26 E + angle 251° 06' 328° 29' - 180° Bearing T.M. 86 - T.M .87 148° 29' + 04' 148° 33' S31°27 E Fixed bearing T.M. 86- T.M.87148°33' .-. Error 04' Horizontal length T.M. 64-1(1 in 6 = 9° 28') = 286*1 cos 9° 28' = 282'2 ft Co-ordinates Ae Se Ae' Line Length 64-1 282-2 1-2 273-2 2-86 146-4 Bearing S 70°14'E S 33°16'E N 77°26'E + 265-6 - 0-5 + 265-1 + 153-2 - 0-5 + 152-7 + 142-9 - 0-2 + 142-7 + 561-7 - 1-2 An Sn An' - 95-4 -0-5 - 95-9 - 233-5 - 0-4 - 233-9 + 31-9 - 0-2 + 31-7 + 31-9 - 1-1 - 328-9 - 297-0 From the theodolite station co-ordinates, E N T.M. 64 45603-1 35 709*9 T.M. 86 46163-6 35 411-8 AE + 560-5 AN - 298-1 '• Error in traverse = E + 1*2, N + 1*1 TRAVERSE SURVEYS 323 Applying Bowditch's method (see p. 330), a-, 1*2 x length 1*2 x / 2 length 701-8 - 1/1 X 1U SN 1-1 x / 2/ 1-1 X 701-8 - = 1-57 x 10~ 3 Total co-ordinates (adjusted) E N T.M.64 45603-1 35709-9 1 45868-2 35614-0 2 46020-9 35380-1 X /. 86 46163-6 35411-8 6.52 Traverses which return to their starting point The closing error may be expressed as either (a) the length and bearing of the closing line or (b) the errors in latitude and departure. To make the traverse consistent, the error must be distributed and this can be done by adjusting either (a) the lengths only, without altering the bearings or (b) the length and bearing of each line by adjusting the co-ordinates. 6.53 Adjusting the lengths without altering the bearings Where all the angles in a closed traverse have been measured, the closing angular error may be distributed either (a) equally or (b) by weight inversely proportional to the square of the probable error. It may therefore be assumed that the most probable values for the bearings have been obtained and that any subsequent error relates to the lengths, i.e. a similar figure should be obtained. Three methods are proposed: (1) Scale factor axis method. (2) xy method (Ormsby method). (3) Crandal's method. In the following description of these methods, to simplify the solu- tion, the normal co-ordinate notation will be altered as follows: partial departure AE becomes d with error in departure Sd. The sum of the errors in departure XSd becomes Ad. Similarly, partial latitude AN becomes / with error in latitude 81. The sum of the errors in latitude 25/ becomes A/. (1) Scale factor axis method (after R.E. Middleton and Q. Chadwick). This follows the principles proposed for traverses closed on fixed points. 324 SURVEYING PROBLEMS AND SOLUTIONS Graphical solution (Fig. 6. 16) The traverse is plotted and the closing error obtained. It is intended that this error produced should divide the figure into two approximately equal parts. To decide on the position of this line, the closing error bearing is drawn through each station until the above condition is obtained. (a) Original plot with lines drawn parade to closing error A A (b) Adjusted figure \^/ Fig. 6.16 Scale factor axis method The lines above XD need to be reduced by a scale factor so as to finish at D 2 midway between D and D,. The lines below XD need to be enlarged by a scale factor so as to finish at D 2 . TRAVERSE SURVEYS 325 Example 6.8 (Fig. 6.17) Fig. 6.17 Graphical adjustment (1) The traverse is plotted as ABCDEA^ -closing error AA y . (2) The closing error is transferred to station D, so that the traverse now plots as D^E^ABCD. D^D 2 is produced to cut the traverse into two parts at X on line AB. (3) From X, rays are drawn through B, C, D, £, and A. (4) From 0, midway between D, and D, lines are drawn parallel to D,/},, giving 0E 2 , and parallel to DC, giving 0C 2 . Lines paral- lel to BC and EA^ through C 2 and E 2 give B z and A 2 . The figure A 2 B 2 C 2 0E 2 A 2 is the adjusted shape with the bear- ings of the lines unaltered, i.e. OX Lines XB, BC and CD are reduced in length in the ratio jrzs OX Lines XA, AE and ED y are enlarged in length in the ratio pr~- (5) If the traverse is to be plotted relative to the original station A, then all the new stations will require adjustment in length and bear- ing A Z A, giving B 3 , C 3 , D 3 , E 3 . The final figure is AB 3 C 3 D 3 E A. 326 SURVEYING PROBLEMS AND SOLUTIONS The graphical solution can now be expressed mathematically as follows (after R.E. Middleton and O. Chadwick). In the figure XBCD, Fig. 6.18, the lengths need to be reduced by a scale factor 1 XD As the co-ordinates are dependent on their respective lengths, the par- tial co-ordinates can be multiplied by the factor k v Let the partial departure be d and the partial latitude /, the error in d be 8d and the error in / be 81. Expressing this as a correction, 8d = d -dk, = d(l - fc,) XD-XD d- XD D,D = d^- XD _ partial departure x terror length of axis of error Thus for all lines above the line XD, dd = Ad traverse error (£),D) where Also X = 2 x axis of error (XD) 81 = XI (6.13) (6.14) (6.15) Similarly in figure XD^E^A the lengths need to be enlarged by a scale factor 8d XD 2 ~xb~. = d(k z -\) = d ( XD * - XD ) = d. XD, D£, XD, TRAVERSE SURVEYS 327 Thus similarly for all lines below line XD. 8d = fxd traverse error (DDi) (6.16) where M 2 x axis of error (XD t ) Also 81 = fil (6.17) By comparison with the traverse lines DD, is small and thus XD, ~ XD 21 XD 2 A ~ /z Summary Sd = Ad 5/ = A/ The sign of the correction depends on the position of the line, i.e. whether the line needs to be reduced or enlarged. N.B. A special case needs to be dealt with, viz. the line AB inter- sected by line DD, produced to X. XB must be reduced AX must be enlarged. XB 8d XB 8d AX AX = + V d AB X^g 8d l AB = fid l AB Also °MB = V-IaB AX-XB AB AX- XB AB (6.18) (6.19) (2) xy (Ormsby) method (Fig. 6.19) If any line is varied in length by a fractional value then the partial co-ordinates will be varied in the same proportion without altering the bearing, i.e. 8s _ 8d_ _ 8l_ s ~ d ~ T Let (a) the lines in the NE/SW quad- rants be altered by a factor x and the lines in the NW/SE quadrants by a factor y, Fig. 6. 19 328 SURVEYING PROBLEMS AND SOLUTIONS (b) the sign of all the terms in summation of the partial co-ordin- ates in one of the equations (say Eq.6.20) be the same as the sign of the greater closing error, (c) the sign in the other equation (say Eq.6.21) be made consis- tent with the figure, to bring the corrections back to the same bearing (Fig. 6. 20). r~ — "71 i i \j- i i s + Ad>AI + a/ i j / \Al I / / + + i.e. Fig. 6.20 + Ad = xd x + yd z + xd z + yd 4 + Ad = x(d, + d 3 ) + y(d 2 + d 4 ) (6.20) + A/ = x/, - y/ 2 + xl 3 - y/ 4 . i.e. + A/ = x(/, + / 3 ) - y(/ 2 + / 4 ) where the partial co-ordinates are d, /,; d 2 l 2 , etc. (6.21) The solution of these simultaneous equations gives values for x and y which are then applied to each value in turn to give the correc- ted values of the partial co-ordinates. (3) CrandaVs method by applying the principle of least squares, Fig. 6.21. Let the length of each side be varied by a fraction x of its lengths, then if / and d be the partial latitude and departure of the line AB of bearing 6, they also will be varied by xl and xd respectively. If the probable error in length is assumed to be proportional to V s ' then the weight to be applied to each line will be 1/s, i.e. TRAVERSE SURVEYS 329 wt oc 1 (probable error) 2 P.E. OC y/J wt oc — s (The value of x is thus depen- dent on the length of the line.) By the theory of 'Least Squares', the sum of the weighted residual errors should be a minimum. i.e. 2 x*s* = 2 x s = minimum. Let Fig. 6.21 Ixl = 25/ = A/, total error in latitude. ?Lxd = 18d = Ad, total error in departure. Then differentiating these equations and equating them to zero will give the minima values, i.e. xs 8x = ISx = d8x = (1) (2) (3) The differentiated equations (2) and (3), being conditional equa- tions, should now be multiplied by factors (correlatives) -&, and -k. (Reference: Rainsford Survey Adjustments and Least Squares). Adding all three equations together and equating the coefficients of each 8x to zero, we have 8x(x, s, - fc, /, - k 2 d,) = 8x(x 2 s 2 - /c, l 2 - k 2 d 2 ) = etc. Thus fc,/, + fcjjd, x, = fc,/ 2 + k 2 d 2 etc. Substituting the values of x into the original equations (2) and '3), we have h f * f/| +/c2f M + / / M2 +k 2 d 2 \ + ... = A/ 330 SURVEYING PROBLEMS AND SOLUTIONS i.e. fc,£— +k 2 2— =M (6.22) aiso d, (*■'■ * M.J + da ^ilhh^ + ... . Ad i.e. *, 2 — + * 2 2- = Ad (6.23) Solving Eqs.(6.22) and (6.23), we obtain values for fc, and k 2 . The corrections to the partial co-ordinates then become 81 - k ^ / ,dl (6.25) s 2 s etc. (6.24) 8d, = fc, Ai + ^ 2 i etc s ' s It should be noted that a check on the equations is given by _5__L_JLl_ _ UMi + MI _A_ fe AA A; ^1 d '^ ! Zf + ^^ dl 1 s + 2 s i.e. no change in bearing. The assumption that the probable error in length is proportional to \Js applies to compensating errors. It has been shown that, where the accuracy of the linear measurement decreases, the probable error in length becomes proportional to the length itself, i.e. P.E. oc s 1 i.e. wt oc — r s 2 The effect of this on the foregoing equations is to remove the factor s from them, i.e. 81, = /c ] ^+^/ 1 d 1 etc (6.26) Sd, = k^d y + k 2 d* etc (6.27) 6.54 Adjustment to the length and bearing Three methods are compared: (1) Bowditch, (2) Transit (Wilson's method), (3) Smirnoff. (1) The Bowditch method (Fig. 6.22). This method is more widely used than any other because of its simplicity. It was originally devi- sed for the adjustment of compass traverses. TRAVERSE SURVEYS 331 Bowditch assumed that (a) the linear errors were compensating and thus the probable error (P.E.) was proportional to the square root of the distance s, i.e. P.E. oc y^- and (b) the angular error 86 in the d would produce an equal displace- ment B, B 2 at right angles to the line AB. A resultant AB 2 is thus develo- ped with the total probable error. BB Z = Vfi.S 2 + B,Bi = y/2B^B (BtB = B,B 2 ) Fig. 6.22 also = y/Sl 2 + Sd 2 where 81 and 8d are the corrections to the partial co-ordinates. The weight, as before, becomes 1/s. By the theory of least squares s («' + "*) ■ a minimum The conditional equations are: 181 = A/ (1) 28d = Ad (2) As in the previous method, using correlatives, differentiation of each equation gives: ^ 81 8(8Q 8d 8(8d) 2* + = s s 28(81) = 2 8(8d) = Multiplying the last two equations by the correlatives -fc, and -k z respectively, adding the equations and equating the coefficients of 5(50 and 5(5d) to zero we have: 5(5/) [H ■ i.e. 5/, = s,*, dl z = s 2 k, etc. 5(5d)|— ~K\ = i.e. 5d, = 5,^ Substituting the values into equations (1) and (2), 8d z = s z k. etc. 332 SURVEYING PROBLEMS AND SOLUTIONS A/ fc, 2 s = A/ i.e. k x = ■=- k z 2s = Ad i.e. k 2 = The corrections to the latitudes 67, = s 2s Ad Is A/_ 2s (6.28) Ad sL = s, =— - etc. ^ 2s Ad to the departures od, = s. -=— 2s Ad od, = s, ■=— etc. 2s i.e. Correction to the partial co-ordinate = total correction length of corresponding side x ______________ _ ___ __ ______ total length of traverse The effect at a station is that the resultant BB 2 will be equal to the closing error x s/2s and parallel to the bearing of the closing error, -1 -d i.e. through an angle a = tan — , Sd = tan" 1 — 5/ The total movement of each station is therefore parallel to the closing error and equal to 2 (lengths up to that point) , . x closing error. total length of traverse The correction can thus be applied either graphically in the manner originally intended or mathematically to the co-ordinates. Jameson points out that the bearings of all the lines are altered unless they lie in the direction of the closing error and that the maxi- mum alteration in the bearing occurs when the line is at right angles to the closing bearing, when it becomes 2s x closing error closing error 86 rad 2s The closing error expressed as a fraction of the length of the tra- verse may vary from 1/1000 to 1/10000, so taking the maximum error as 1/1000 TRAVERSE SURVEYS 333 86" = 2 ° 6265 = 206" 1000 = 03' 26" a value far in excess of any theodolite station error. A change of bea- ring of 20" represents 1/10000 and this would be excessive even using a 20" theodolite. Graphical Solution by the Bowditch Method (Fig. 6.23) Fig. 6.23 (1) Plot the survey and obtain the closing error AA E . (2) Draw a line representing the length of each line of the traverse to any convenient scale. (3) At A E draw a perpendicular A E A^ , equal to the closing error and to the same scale as the plan. (4) Join AA X , forming a triangle AA X A E , and then through B, C and D similarly draw perpendiculars to cut the line AA y at B„ C, and D, . (5) Draw a line through each station parallel to the closing error 334 SURVEYING PROBLEMS AND SOLUTIONS and plot lines equal to BB U CC X and DD X , giving the new figure Afi,C,D,A . (2) The Transit or Wilson method. This is an empirical method which can only be justified on the basis that (a) it is simple to ope- rate, (b) it has generally less effect on the bearings than the Bowditch method. It can be stated as: the correction to the partial co-ordinate ., .. f ,. A closing error in the co-ordinate = the partial co-ordinate x laie x — Z partial co-ordinates (ignoring the signs) (6.29) (6.30) 8d < - d ' Td (6.31) (3) The Smirnoff method. The partial latitude / of a line length s and bearing 6 is given by / = s COS0. If the two variables s and cos0 are subjected to errors of 8s and 8(cosd} respectively, then 1 + 81 = (s + 8s)[cos0 + S(cos0)] Subtracting the value of / from each side and neglecting the small value 8s S(cos0), gives 81 = 8s cos# + s 8(cosd) Dividing both sides by /, — _ 8s cos e s _S(cos0) / s cos0 s cos 6 i.e. 81 8s 8(cosd) j = — + , ' (6.32) / s cos# i.e. the relative accuracy in latitude = the relative accuracy in distance + the relative accuracy in the cosine of the bearing Thus, in a traverse of n lines, M 1 8s * j. i g ( cos fl) s, cos 6/, Sl 2 = I — + l ——— — etc. s 2 - cosu 2 TRAVERSE SURVEYS 335 25/ = A/ (total error in latitude) v 8s SCcosfl, ) S(cos0 2 ) 5(cos0 n ) — 2*1 + I. — + / 2 — — + ... l n s cos0, " cos0 2 cos0 n Similarly, as d = s sin 0, — = £f S(sinfl) d s sin (6.33) 18d = Ad = 2d«f + d , S(5in ^ ) + w g(sin ^> + .. <* 5 < sin ^ s sin0, 2 sin0 2 n sin0 n where the linear relative accuracy 8s/ s is considered constant for all lines. From the above equations, S(cos0)l 8s _ 1_[ s ~ 2/1 ^ 1>-Sd«l (6.35) s zdl sin0 J The value of 8s/ s should closely approximate to the actual acc- uracy in linear measurement attained if the traverse consists of a large number of lines, but in short traverses there may be quite a large disc- repency. In such cases the ratio shows the accuracy attained as it affects the closing error. The ratio is first worked out separately for latitude and departure from Eqs. 6.34/6.35 and these allow subsequent corrections to be applied as in Eqs. 6.32/6.33. N.B. The precision ratios for cosine and sine of the bearings are obtained by extraction from trigonometrical tables. Special attention is necessary when values of 0° or 90° are involved as the trigonomet- rical values of oo will be obtained. The traverse containing such bear- ings may be rotated before adjustment and then re-orientated to the original bearings. To calculate the precision ratio for cos 0, Let = 60° + 6" cos0 = 0-5 8 (cos 0) difference/ 6" = 0-000025 S(cos0) _ 0-000025 1 cos0 ~ 0^5 = 20000 The ratio is therefore proportional to the angular accuracy. Where a negative ratio 8s/ s is obtained it implies that the angular precision has been over-estimated. 336 SURVEYING PROBLEMS AND SOLUTIONS As the precision of the angular values increases relative to the linear values, the precision ratios of the former reduce and become negligible. 8s = Al s 11 Thus and, substituting this into the partial latitude equation, A/ 81, = Z, 21 Similarly, sd, - d, Td These equations thus reduce to method of adjustment (2), the Transit Rule. 6.55 Comparison of methods of adjustment Example 6.8 Line AB BC CD DE EF FA I Bearing 6 Length s + - + - 045° 00' 514-63 363-898 363-898 090° 00' 341-36 341-360 0-0 180° 00' 324-15 0-0 324-150 210° 00' 462-37 231-185 400-420 300° 00' 386-44 334-667 193-220 320° 16' 217-42 138-978 167-202 2s 2246-37 705-258 704-830 704-830 1 724-320 724-570 724-320 Ad + 0-428 A/ -0-250 Assuming the co-ordinates of A (0,0) Total Co-ordinates D 0-0 + 363-898 + 705-258 + 705-258 + 474-073 + 139-406 + 0-428 Bearing of closing line AA, = tan-' +0-428/-0-250 = S59°43'F Length of closing line A A, = 0-428 cosec59°43' = 0-496 A B C D E F A, L 0-0 + 363-898 + 363-898 + 39-748 -360-672 -167-452 - 0-250 TRAVERSE SURVEYS 337 (1) 'Axis' scale factor method (Figs. 6.24 and 6.25) 400r 300 200 700 800 -100- •200- -300- -400 Fig. 6.24 i Lines above axis \D ! require reduction Lines below axis \io«>2> require enlargement / / ^*\ ^i 'E E, ^Axis Fig. 6.25 Scaled values from plotting (station D is chosen to contain the closing error as the axis approximately bisects the figure): DX = 487 AX ' = 406 XB = 108 From Eqs. 6.14/6.15, 8d = \d 81 = \l AA X 0-496 where A = 8d 2 x Axis(DX) 2x487 AX - XB = 5-1 x 10" 4 AB = AB x Xd AB 406 - 108 c.-i 1n -4 o*o o = — ■=—: x + 5*1 x 10 x 363-9 514 = +0-105 338 SURVEYING PROBLEMS AND SOLUTIONS (AX requires enlarging, AX > XB) 8d BC = -5-1 x 10" 4 x +341-4 \ = -0-174 SdcD = 0-0 8d DE = +5-1 x 10 -4 x -231-2 = -0-118 8d EF = +5-1 x 10 -4 x -334-7 = -0-171 8d FA = 81ab = $Ibc f5-l x 10" 4 x -139-C = -0- 070 105 . s.i in -4 406 - f5-l x 10 X r—- A 514 108 x +363-9 - +0- = 0- 8l CD = -5-1 x 10" 4 x -324-2 = +0-165 8l DE = +5-1 x 10" 4 x -400-4 [ = -0-204 8l EF = +5-1 x 10" 4 x +193-2 _ = +0-098 8l FA = +5-1 x 10 -4 x +167-2 = +0-086 Co-ordinates d 8d d n la 81 AB + 363-898 +0-105 + 364-063 + 363-898 + 0-105 + 364-063 BC + 341-360 -0-174 + 341-186 0-0 0-0 0-0 CD 0-0 0-0 0-0 -324-150 +0-165 -323-985 DE -231-185 -0-118 -231-303 -400-420 -0-204 -400-624 EF -334-667 -0-171 -334-838 + 193-220 +0-098 + 193-318 FA -138-978 -0-070 -139-048 + 167-202 + 0-086 + 167-288 -0-533 +0-454 +0-105 -0-204 Ad -0-428 M +0-250 (2) Ormsby's xy method (Fig. 6.26) Bearing Term d 8d / 81 AB 045° 00" X + 363-898 -0-063 +363-898 -0-063 BC 090° 00' X + 341-360 -0-060 0-0 0-0 CD 180° 00' y 0-0 0-0 -324-150 +0-181 DE 210° 00' X -231-185 -0-040 -400-420 -0-069 EF 300° 00' y -334-667 -0-187 + 193-220 + 0-108 FA 320° 16' y -138-978 -0-078 + 167-202 +0-093 Ad -0-428 +0-382 -0-132 A I +0-250 TRAVERSE SURVEYS 339 N.B. Term x is assumed to be 0° -» 90° inclusive y is assumed to be 90°-* 180° As the error in departure is greater, the equation takes the sign of the correction, i.e. -364 x -341 x - Oy -231 x -335y -139y = -0-428 (1) (AB) (SC) (CD)(DE) (£F) (FA) Adjusting the latitude values and signs to make them consistent, -364 x + Ox + 324y -400 x +193y + 167y = +0-250 (2) Fig. 6.26 Simplifying the equations, -936x -474y = -0-428 (1) -764* +684y = +0-250 (2) Solving the equations simultaneously, x = +1-74 x HT 4 y = +5-59 x 10~ 4 The values of x and y are now applied to each term to give corrections as above. (3) CrandaVs method (least squares) (a) Probable error oc length s Using equations, k,2ld + k 2 ^d z k,Xl 2 + k 2 lid Ad A/ 340 SURVEYING PROBLEMS AND SOLUTIONS AB EC CD DE EF FA d + 363-898 + 341-360 0-0 -231-185 -334-667 -138-978 Ad -0-428 * + 363-898 0-0 -324-150 -400-420 + 193-220 + 167-202 M +0-250 + 132421-8 116526-6 f 0-0 53 446-5 112002-0 19 314-9 433 711-8 (Id 2 ) Id + 132 421-8 0-0 0-0 + 92571-1 - 64664-4 - 23237-4 + 224 992-9 - 87901-8 + 137091-1 (2/d) 137091-U,+ 433711-8fc 2 = -0-428 463 121-7 fc,+ 137 091-1 k 2 = +0-250 Solving simultaneously, fc, - +9-1768 x 10~ 7 k 2 = -1-2769 x 10~ 6 Substituting in the equations /c, /, d, + k 2 d 2 = 5d, k,/ t + fc 2 /,d, = S/, gives the correction for each partial co-ordinate: + 132421-8 0-0 105 073-2 160 336-2 37 334-0 27956-5 463121-7 (1) (2) kyld k 2 d z Sd M 2 k 2 ld 81 AB + 0-122 -0-169 -0-047 + 0-122 -0-169 -0-047 EC 0-0 -0-149 -0-149 0-0 0-0 0-0 CD 0-0 0-0 0-0 + 0-096 0-0 + 0-096 DE + 0-085 -0-068 + 0-017 + 0-147 -0-118 +0-029 EF -0-059 -0-143 -0-202 + 0-034 + 0-083 +0-117 FA -0-021 -0-026 -0-047 +0-026 f 0-029 + 0-055 + 0-207 -0-555 -0-445 + 0-425 - 0-287 +0-297 -0-080 +0-127 -0-428 + 0-017 -0-175 + 0-250 + 0-112 -0-047 + 0-127 -0-428 -0-175 + 0-250 TRAVERSE SURVEYS 341 (b) Probable error oc -y/ s AB BC CD DB EF FA s 514-63 341-36 324-15 462-37 386-44 217-42 1/s 0-001 945 0-002929 0-003 085 0-002 163 0-002 588 0-004 599 d7s 257-295 341-306 0-0 115-605 289-861 88-292 Id/s + 257-295 0-0 0-0 + 200-231 -167-351 -106-869 17 s 257-295 0-0 324-151 346-807 96-620 128-572 2d 2 /s 1092-359 + 457-526 S/ 2 /s 1153-445 -274-220 Xld/s +183-306 Using equations s s k, 2— + k s s ld = Ad M and substituting values gives + 183-306 k, + 1092-359 k 2 = + 1153-445 k, + 183-306 k 2 = Solving simultaneously, *, = +2-867 x 10~ 4 k 2 = -4-398 x 10~ 5 Substituting values into equations -0-428 + 0-250 = 8d, 8d -0-039 -0-150 0-0 + 0-006 -0-175 -0-070 -0-434 +0-006 +0-052 -0-428 -0-428 /c, Id/s k 2 d z /s AB + 0-074 -0-113 BC 0-0 -0-150 CD 0-0 0-0 DE + 0-057 -0-051 EF -0-048 -0-127 FA -0-031 -0-039 + 0-131 -0-480 -0-079 + 0-052 *, 1+ K /c,/ 2 /s + 0-074 0-0 + 0-093 + 0-099 + 0-027 + 0-037 + 0-330 -0-080 + 0-250 (1) (2) AA = 8L k 2 ld/s -0-113 0-0 0-0 -0-088 + 0-074 + 0-047 -0-201 + 0-121 -0-080 81 -0-039 0-0 + 0-093 + 0-011 + 0-101 + 0-084 + 0-289 -0-039 + 0-250 342 SURVEYING PROBLEMS AND SOLUTIONS (4) Bowditch's method s d I 8d 81 AB 514-63 +363-898 +363-898 -0-098 +0-057 BC 341-36 +341-360 0-0 -0-065 +0-038 CD 324-15 0-0 -324-150 -0-062 +0-036 DE 462-37 -231-185 -400-420 -0-088 +0-052 EF 386-44 -334-667 +193-220 -0-074 +0-043 FA 217-42 -138-978 +167-202 -0-041 +0-024 Is 2246-37 -0-428 +0-250 [id -0-428 A/ +0-250 Xd 1410-088 2/ 1448-890 Using the formulae, 8d =^xs = -°;f* 5 = -1-906 x 10~ 4 5 2s 2246-37 81 = *L s = +°' 250s = +1-112 x 10- 4 s 2s 2246-37 Ex. Sd, = -1-906 x 10" 4 x 514-63 = -0-098 81, = +1-112 x 10~ 4 x 514-63 = +0-057 (5) Transit or Wilson's method Using the formulae, 8d = M. x d = -°' 428<i = -3-035 x 10" 4 d 2d 1410-088 5/ = A/ z = +0-250 / = +1 . 725 10 -4 Z 2/ 1448-890 Ex. Sd, = -3-035 x 10 -4 x +363-898 = -0-111 5/, = +1-725 x 10~ 4 x +363-898 = +0-063 8d 81 AB -0-111 + 0-063 BC -0-103 0-0 CD -0-0 + 0-056 DE -0-070 + 0-069 EF -0-102 +0-033 FA -0-042 +0-029 Ad -0-428 A/ +0-250 TRAVERSE SURVEYS 343 (6) Smirnoff's method N.B. As bearings of BC and CD are 90° and 180° respectively, the values of g ^ cos90 > and 5 < sinl80 > will be infinity, cos 90 sin 180 Thus the whole survey is turned clockwise through 20° giving new bearings (with an accuracy of ± 10") e s AB 065-00 514-630 BC 110-00 341-360 CD 200-00 324-150 DE 230-00 462-370 EF 320-00 386-440 FA 340-16 217-420 sin0 8 (sin xlO" 6) -6 ^ dS(sin#) sin# s 8d d (Adj) AB 0-906 308 20 + 466-413 +0-010 0-074 -0-084 + 466-329 BC 0-939693 17 -t 320-774 +0-006 0-051 -0-057 +320-717 CD 0-342020 45 110-866 -0-013 0-017 -0-030 -110-896 DE 0-766044 31 - 354-196 -0-014 0-056 -0-070 -354-266 EF 0-642 788 37 248-399 -0-014 0-039 -0-053 -248-452 FA 0-337643 46 + 73-410 -0-010 787-187 786-871 0-012 -0-022 - 73-432 Ad + 0-316 0-067 0-249 - 0-316 2d 1574-058 cosO 5 ( cos i ) / l8(cos&) z Ss_ Sl /(Adj) xlO" 6 cos e AB 0-422618 44 +217-492 0-023 0-046 +0-069 +217-561 BC 0-342020 45 -116-752 0-015 0-025 +0-040 -116-712 CD 0-939693 17 -304-601 0-006 0-065 +0-071 -304-530 DE 0-642788 37 -297-206 0-017 0-064 +0-081 -297-125 EF 0-766 044 31 +296-030 0-012 0-064 +0*076 +296-106 FA 0-941274 17 +204-652 0-004 0-044 +0-048 +204-700 + 718-174 -718-559 A/ - 0-385 0-077 +0-385 21 1436-733 344 SURVEYING PROBLEMS AND SOLUTIONS Departure 8s = J_ M d8(sin0) ] sintf J 1 [+0-316-0-067] = °' 249 1574-1 1574-1 = 1-581 x 10~ 4 Latitude 8s = 1T A/ /a(cosg) 1 s £/[ cos0 J = — L_ [0-385 - 0-077] = -°-^ - 8 - = 2-14 x 10~ 4 1436-7 1436-7 The co-ordinates must now be transposed to their original orienta- tion. Using Eqs. (3.29/30) x 2 = x, cos 20 - y, sin 20° y 2 = y, cos 20 + x, sin 20° cos 20° = 0-939 693 sin 20° = 0-342020 Line AB x, +466-329 + 438-206 - 74-410 = +363-796 8d = -0-102 y, +217-561 + 204-441 + 159-494 = +363-935 81 = +0-037 Line BC x, +320-717 + 301-376 + 39-918 = +341-294 8d = -0-074 y, +116-712 -109-673 + 109-692 = + 0-019 81 = + 0-019 Line CD x, -110-869 -104-208 + 104-155 = - 0-053 8d = -0-053 y, =304-530 -286-165 - 37-929 = -324-094 5/ = + 0-056 Line DE x, -354-266 -332-901 + 101-623 = -231-278 8d = -0-093 y, -297-125 -297-206 - 121-166 = -400-372 81 = -0-048 Line EF x, -248-452 -233-469 - 101-274 = -334-743 8d = -0-076 y, +296-106 + 278-249 - 84-976 = +193-273 81 = +0-053 Line FA x, - 73-432 - 69-004 - 70-011 = -139-015 8d = -0-037 y, +204-700 + 192-355 - 25-115 = +167-240 81 = + 0-038 TRAVERSE SURVEYS Analysis of corrections (Figs. 6.27 — 6.31) 345 Line AB fs- 514ft J 045* 00; Fig. 6. 27 thod 86 8s 8d 5/ 1 +0-15 +0-105 +0-105 2 -0-09 -0-063 -0-063 3a -0-07 -0-047 -0-047 3b -0-05 -0-039 -0-039 4 -44" -0-03 -0-098 +0-057 5 -48" -0-03 -0-111 + 0-063 6 -36" -0-05 -0-102 + 0-037 Line BC (S-3AM 090* 00' « 4 , 3o o- o— — L o — _ — • > 1 3b 6d Fig. 6.28 Method 86 8s 8d 81 1 -0-17 -0-174 0-0 2 -0-06 -0-060 0-0 3a -0-15 -0-149 0-0 3b -0-15 -0-150 0-0 4 + 22" -0-07 -0-065 +0-038 5 -0-10 -0-103 0-0 6 + 11" -0-07 -0-074 +0-019 346 SURVEYING PROBLEMS AND SOLUTIONS Line CD (S=-324) 180* 00' 6 4o- 1 Fig. 6. 29 Method 86 8s 8d 81 1 +0-17 0-0 + 0-165 2 +0-18 0-0 +0-181 3a +0-10 0-0 +0-096 3b +0-09 0-0 + 0-093 4 + 44" 0-07 -0-062 +0-036 5 +0-06 0-0 +0-056 6 + 38" 0-06 -0-053 +0-056 Line Of s = 462 6/ 210* 00' ■0-10 / /D 5 4 V v -oio ) *\b ♦0-10 60 V / ^r 2 9C ' to closing error Fig. 6. 30 Method 86 8s 8d 81 1 0-23 -0-118 -0-204 2 0-08 -0-040 -0-069 3a 0-04 +0-017 +0-029 3b 0-02 + 0-006 +0-011 TRAVERSE SURVEYS 347 (N.B. 90 thod 86 8s 8d 81 4 +45" 0-0 -0-088 +0-052 5 +44" 0-03 -0-070 +0-069 6 + 26" 0-09 -0-093 -0-048 ° to closing error) Line EF s = 386 300° 00' \3o 61 3?\? \4 ♦01 5/> ■ F i -0-1 ♦ 0-1 6d On line of closing error Method 86 8s 8d 81 1 0-19 -0-171 +0-098 2 0-12 -0-187 +0-108 3a 0-23 -0-202 +0-117 3b 0-19 - 0-175 +0-101 4 0-09 - 0-074 + 0-043 5 -5" 0-10 -0-102 +0-033 6 0-09 -0-076 + 0-053 (N.B. On line of closing error) The following may be conjectured: (1) The first four methods do not change the bearings. (2) Method (1) has a greater effect on the linear values than any other. (3) There is no change in bearing in Wilson's method when the line coincides with the axes. (4) There is little or no change in bearing on any line parallel to the closing error in any of the methods analysed - maximum linear correc- tion. (5) Wilson's method has less effect on the bearings than Bowditch's, but more than Smirnoff's. (6) The maximum change in bearing occurs at 90° to the closing error— maximum linear correction. 348 SURVEYING PROBLEMS AND SOLUTIONS Exercises 6(c) (Traverse Adjustment) 12. The mean observed internal angles and measured sides of a closed traverse ABCDA (in anticlockwise order) are as follows: Angle Observed Value Side Measured Length (ft) DAB 97° 41' AB 221-1 ABC 99° 53' BC 583-4 BCD 72° 23' CD 399-7 CD A 89° 59' DA 521-0 Adjust the angles, compute the latitudes and departures assuming that D is due N of A, adjust the traverse by the Bowditch method; and give the co-ordinates of 3, C and D relative to A. Assess the accuracy of these observations and justify your assess- ment. (I.C.E. Ans. B -30-3 N, + 219-7 E, C +523-9 N, +397-5 E, D +522-6 N, - 1-2 E) 13. The measured lengths and reduced bearings of a closed theodo- lite traverse ABCD are as follows: Line Length (ft) Bearing AB 454-9 Due N BC 527-3 Due W CD 681-0 S25°18'W DA 831-2 N78°54'E (a) Adjust the traverse by the Bowditch method and taking A as the origin, find the co-ordinates of B, C and D. (b) Assess the accuracy of the unadjusted traverse. (c) Suggest, and outline briefly, an alternative method of adjusting the traverse so that the bearing of AB is unaltered by the adjustment. (I.C.E. Ans. B 455-0 N, 0-5 E, C 455-1 N, 526-2 W, D 160-3 S, 816-5 W) 14. The following lengths, latitudes and departures refer to a closed traverse ABCDEA : Length Latitude Departure AB 3425-9 3425-9 BC 938-2 812-6 469-1 CD 4573-4 2287-1 -3961-0 DE 2651-3 -2295-7 -1325-9 EA 1606-4 - 803-0 1391-1 Adjust the traverse by the Bowditch method, finding the corrected latitudes and departures to the nearest 0*1 ft. TRAVERSE SURVEYS 349 Discuss the merits and demerits of this method, with particular reference to its effect on lines CD and DE. (L.U. Ans. AB - 0*3, + 3426-1 BC + 812-5, + 469-2 CD + 2286-8, -3960-7 DE -2295-9, -1325-8 EA - 803-1, + 1391-2) 15. In a closed traverse ABCDEFA the lengths, latitudes and departures of lines (in ft) are as follows: Line AB BC CD DE EF FA Length 1342-0 860-4 916-3 1004-1 1100-0 977-3 Latitude -1342-00 -135-58 +910-35 +529-11+525-99-483-23 Departure 0-0 +849-65 +104-26 + 853-38- 966-08-849-42 Adjust the traverse by the Bowditch method and give the corrected co-ordinates of A as (0, 0) (L.U. Ans. A 0-0, 0-0 D -569-56, +958-04 B -1343-00, +1-77 E - 41-20, +1812-74 C -1479-22, + 852-56 F +483-96, + 848-12 16. A traverse ACDB is surveyed by theodolite and chain. The lengths and quadrantal bearings of the lines, AC, CD and DB are given below. If the co-ordinates of A are x = 0, y = and those of B are x = 0, y = +897-05, adjust the traverse and determine the co-ordinates of C and D. The co-ordinates of A and B must not be altered. Line AC CD DB Length 480-6 292-0 448-1 Bearing N25°19'E N37°53'E N59°00'W (L.U. Ans. C +205-2, 435-0 D +384-4, 665-8) 17. The lengths, latitudes and departures of the lines of a closed traverse are given below. In one of the lines it appears that a chainage has been misread by 40 ft. Select the line in which the error is most likely to have occurred, correct it and adjust the latitudes and departures by the Bowditch method to the nearest 0-1 ft. Line Length (ft) Latitude Departure AB 310-5 + 301-2 + 75-4 BC 695-8 + 267-1 -642-5 CD 492-8 -299-8 -391-1 Line Length (ft) Latitude DE 431-7 -359-1 EF 343-1 + 173-5 FA 401-9 - 49-0 350 SURVEYING PROBLEMS AND SOLUTIONS Departure + 239-6 + 296-0 + 398-9 (L.U. Ans. Line DE 40 ft too long AB +301-1 + 75-6 DE -392-5 +262-0 BC +267-0 -642-1 EF +173-4 +296-2 CD -299-9 -390-8 FA - 49-1 +399-1) 18. (a) Why is the accuracy of angular measurement so important in a traverse for which a theodolite and steel tape are used? (b) A and D are the terminals of traverse ABCD. Their plane rectangular co-ordinates on the survey grid are: Eastings Northings A +5861-14 ft +3677-90 ft D +6444-46 ft +3327-27 ft The bearings adjusted for angular misclosure and the lengths of the legs are: AB 111° 53' 50" 306-57 ft BC 170° 56' 30" 256-60 ft CD 86° 43' 10" 303-67 ft Calculate the adjusted co-ordinates of B and C (N.U. Ans. B E 6100*70 N 3563-47 C E 6141-19 N 3309-99) 19. From an underground traverse between two shaft-wires, A and D, the following partial co-ordinates in feet were obtained: AB E 150-632 ft S 327-958 ft BC E 528 -314 ft N 82-115 ft CD E 26-075 ft N 428 -862 ft Transform the above partials to give the total Grid co-ordinates of station B given that the Grid co-ordinates of A and D were: A E520 163-462 metres N 432182-684 metres D E 520 378-827 metres N 432238-359 metres (aide memoire : X = *, + k(x-y6) Y = y, + k(y + xd)) (N.R.C.T. Ans. B E 520 209-364 N 432082-480) 20. (a) A traverse to control the survey of a long straight street forms an approximate rectangle of which the long sides, on the pavements, are formed by several legs, each about 300 ft long and the short sides are about 40 ft long; heavy traffic prevents the measurement of lines obliquely across the road. A theodolite reading to 20" and a tape TRAVERSE SURVEYS 351 graduated to 0*01 ft are used and the co-ordinates of the stations are required as accurately as possible. Explain how the short legs in the traverse can reduce the accuracy of results and suggest a procedure in measurement and calculation which will minimize this reduction. (b) A traverse TABP was run between the fixed stations T and P of which the co-ordinates are: E N T +6155-04 +9091-73 P +6349-48 +9385-14 The co-ordinate differences for the traverse legs and the data from which they were calculated are: Length Adjusted Bearing AE AN TA 354-40 210°41'40" -180-91 -304-75 AB 275-82 50° 28' 30" +212-75 +175-54 BP 453-03 20° 59' 50" +162-33 +422-95 Applying the Bowditch rule, calculate the co-ordinates of A and fi. (L.U. Ans. A E 5974-22 N 8786-87 B E 6187-04 N 8962-33) 21. The co-ordinates in feet of survey control stations A and B in a mine are as follows: Station A E 8432-50 N 6981-23 Station B E 9357-56 N 4145-53 Undernoted are azimuths and distances of a traverse survey between A and B. Line Azimuth Horizontal Distance 4-1 151° 54' 20" 564-31 1-2 158° 30' 25" 394-82 2-3 161° 02' 10" 953-65 3-4 168° 15' 00" 540-03 4-fi 170° 03' 50" 548-90 Adjust the traverse on the assumption that the co-ordinates of stations A and B are correct and state the corrected co-ordinates of the traverse station E N (M.Q.B./S Ans. A 8432-50 6981-23 1 8698-24 6483-55 2 8842-91 6116-28 3 9152-83 5214-64 4 9262-82 4686-07 fi 9357-56 4145-53) 352 SURVEYING PROBLEMS AND SOLUTIONS 22. Define the terms 'error of closure' and 'fractional linear closing error' as applied to closed traverse surveys. What error of closure would be acceptable for a main road traverse survey underground ? Starting with the equations / = s cos a d = s sin a derive the Smirnoff equations d JL = lUdL) S Zl{ V/ d(cosa)) cos a j and *? = -Lf [dD) Sd <*(sina)] sin a J where a = bearing angle s = length of traverse draft ds/s = linear precision ratios d(cosa) , d(sina) ana cos a sin a = angular precision ratios 11 = sum of latitudes Xd = sum of departures dh = total closing error in latitudes dD = total closing error in departures. (N.U.) Exercises 6(d) (General) 23. The following are the notes of a theodolite traverse between the faces of two advancing roadways BA and FG, which are to be driven until they meet. Calculate the distance still to be driven in each roadway. Line Azimuth Distance (ft) AB 267° 55' 150 BC 355° 01' 350 CD 001° 41' 315 DE 000° 53' 503 EF 086° 01' 1060 FG 203° 55' 420 (Ans. BA produced 352-6 ft FG produced 916-5 ft) 24. The following measurements were made in a closed traverse, ABCD A = 70° 45' ; D = 39° 15' TRAVERSE SURVEYS 353 AB = 400 ft ; CD = 700 ft ; AD = 1019 ft Calculate the missing measurements. (L.U./E Ans. B = 119° 58', C = 130° 02', BC = 351-1 ft) 25. Particulars of a traverse survey are as follows: _ine Length (ft) Deflection Angle AB 330 B 76° 23' right BC 515 C 118° 29' right CD 500 D 79° 02 'right DA 375 A 86° 06' right Bearing of line AB 97° 15' Prepare a traverse sheet and so calculate the length and bearing of the closing error. (L.U./E Ans. 6*4 ft, N347°04'W) 26. The interior angles of a closed (clockwise) traverse ABCDEA have been measured with a vernier theodolite reading to 20", with re- sults as follows: Angle at A 88° 03' 20" B 117° 41' 40" C 126° 13' 00" D 119° 28' 40" E 88° 35' 00" Adjust the measurements to closure and find the reduced bearings of the other lines if that for line AB is S 42° 57' 20" E. (L.U./E. Ans. BC S48°59'40"W DE N14°54'20"W CD N 68° 41' 40" W EA N 45° 37' 20" E) 27. An approximate compass traverse carried out over marshy ground yielded the following results: Line Length (ft) Bearing AB 386 139° BC 436 50° CD 495 335° DE 271 249° EA 355 200° Plot the traverse to a scale of 100 ft to the inch and adjust it graph- ically to closure. 28. A plot of land is up for sale and there is some doubt about its area. As a quick check, a compass traverse is run along the boundaries. Determine the area enclosed by the traverse from the following data: Line Bearing AB 195° BC 275° CD 182 1 / 2 ° DE 261^° EF 343° FG 5° GH 8oy 2 ° HA 102 1 / 2 ° 354 SURVEYING PROBLEMS AND SOLUTIONS Feet 528 548 813 1293 788 653 1421 778 (I.C.E. Ans. 56 acres) 29. The traverse table below refers to a closed traverse run from station D, through 0,G and H and closing on D. The whole-circle bearing of from D is 06° 26' and G and H lie to the west of the line OD. Compute the latitudes and departures of 0,G and H with reference to D as origin, making any adjustments necessary. Observed Internal Angles Length in feet DO 547-7 OG 939-8 GH 840-2 HD 426-5 Ans. +545-1, + 61-0 G +846-1, -830-7 H +121-6, -408-2) 30. The field results for a closed traverse are: HDO 79° 47' DOG 102° 10' OGH 41° 11' GHD 136° 56' (I.C.E Line Whole Circle Bearing Length (ft) AB 0°00' 166 BC 63° 49' 246 CD 89° 13' 220 DE 160° 55' 202 EF 264° 02' 135 FA 258° 18' 399 The observed values of the included angles check satisfactorily, but there is a mistake in the length of a line. Which length is wrong and by how much ? As the lengths were measured by an accurate 100 ft chain, suggest how the mistake was made. (I.C.E. Ans. Line BC 20 ft short) 31. The following traverse was run from station I to station V be- tween which there occur certain obstacles. TRAVERSE SURVEYS 355 Line Length (ft) Bearing I -II 351-3 N 82°28'E II - HI 149-3 N 30°41'E III - IV 447-3 S 81°43'E IV- V 213-3 S 86°21'E It is required to peg the mid-point of I — V. Calculate the length and bearing of a line from station III to the re- quired point. (I.C.E. Ans. 171-1 ft S42°28'E) 32. Two shafts, A and B, have been accurately connected to the National Grid of the Ordnance Survey and the co-ordinates of the shaft centres, reduced to a local origin, are as follows: Shaft A E 10 055-02 metres N 9768-32 metres Shaft B E 11801-90 metres N 8549-68 metres From shaft A, a connection to an underground survey was made by wires and the grid bearing of a base line was established from which the underground survey was calculated. Recently, owing to a holing through between the collieries, an opportunity arose to make an under- ground traverse survey between the shafts A and B. This survey was based on the grid bearing as established from A by wires, and the co- ordinates of B in relation to A as origin were computed as E 5720-8 ft S 4007 -0 ft Assuming that the underground survey between A and B is correct, state the adjustment required on the underground base line as estab- lished from shaft A to conform to the Nation Grid bearing of that line. (M.Q.B./S Ans. 00°06'30") 33. It is proposed to sink a vertical staple shaft to connect X on a roadway CD on the top horizon at a colliery with a roadway GH on the lower horizon which passes under CD. From the surveys on the two horizons, the undernoted data are available: Upper Horizon Station Horizontal Angle Inclination + 1 in 200 Inclined Length (ft) 854-37 B 276° 15' 45' 88° 19' 10' + 1 in 400 943-21 Remarks co-ordinates of A E 6549 -10 ft N 1356-24 ft Bearing AB N 30° 14' 00" E level 736-21 D 356 SURVEYING PROBLEMS AND SOLUTIONS Lower Horizon Station Horizontal Angle Inclination F G H 193° 46' 45" 83° 03' 10" + 1 in 50 + 1 in 20 level Inclined Length (ft) 326-17 278-66 626-10 Remarks co-ordinates of E E 7704-08 ft N 1210-88 ft Bearing EF N 54° 59' 10" E Calculate the co-ordinates of X (M.Q.B./S Ans. X = E 8005-54 ft N 1918-79 ft) 34. Calculate the co-ordinate values of the stations B,C,D and E of the traverse ABCDEA, the details of which are given below. Data: Co-ordinates of A 1000*0 ft E 1000 -0 ft N Bearing of line AB 0°00' Length of line AB 342*0 ft Interior Angle BAE 27° 18' 00" CBA 194° 18' 40" DCB 146° 16' 00" EDC 47° 27' 20" AED 124° 40' 00" (R.I.C.S. Ans. B 1000-0 E, 1342-0 N C 898-2 E, 1741-1 N D 1035-2 E, 2129-6 N E 1313-4 E, 1607-0 N) 35. The table below gives the forward and back quadrantal bearings of a closed compass traverse. Tabulate the whole-circle bearings corrected for local attraction, indicating clearly your reasons for any corrections. Length (ft) AB 342 BC 412 CD 412 DE 592 EA 683 Line Length (ft) Forward Bearing Back Bearing AB 650 N 55° E S54° W BC 328 S 671/2° E N 66° W CD 325 S 25° W N 25° E DE 280 S 77° W N 75 1 / 2 ° E EA 440 N 64%° W s 63y 2 ° e A gross mistake of 100 ft has been made in the measurement or booking of one of the lines. State which line is in error. Using this corrected length, adjust the departure and latitude of each line of the traverse to close, using Bowditch's method of adjustment. (L.U. Ans. Local attraction at B and E, CD 100 ft too small) TRAVERSE SURVEYS 357 36. It is proposed to extend a straight road AB in the direction AB produced. The centre line of the extension passes through a small farm and in order to obtain the centre line of the road beyond the farm a traverse is run from B to a point C, where A, B and C lie in the same straight line. The following angles and distance were recorded, the angles being measured clockwise from the back to the forward station: ABD = 87° 42' BD = 95-2 ft BDE = 282° 36' DE = 253-1 ft DEC = 291° 06' Calculate (a) the length of the line EC (b) the angle to be measured at C so that the centre line of the road can be extended beyond C. (c) the chainage of C taking the chainage of A as zero and AB = 362 ft. (L.U. Ans. (a) 58-3 ft; (b) 58° 36'; (c) 576-8 ft) 37. The following are the notes of a traverse made to ascertain the position if the point F was in line with BA produced. Line Azimuth Distance AB 355° 30' 600 ft level BC 125° 00' 310 ft rising 1 in 2 CD 210° 18' 378 ft level DE 130° 36' 412 ft level EF 214° 00' 465 ft level Calculate the difference in the azimuths of AF and BA and the extent to which the point F is out of alignment with BA produced. (N.R.C.T. Ans. 0°01'; 0-3 ft) 38. The following notes were made when running a traverse from a station A to a station E: Side W.C. Bearing Length (ft) AB 119° 32' 264-8 BC 171° 28' 162-4 CD 223° 36' 188-3 DE 118° 34' 316-5 A series of levels were also taken along the same route as follows; BS I.S. F.S. R.L. Remarks 6 * 84 246-20 B.M. near A 3-86 Sta. A 11*02 Sta. B 1*32 13-66 C.P. 1 358 SURVEYING PROBLEMS AND SOLUTIONS BS I.S. F.S. R.L. Remarks Sta. C I.S. F.S. 9-66 12-96 13-44 12-88 Sta. D 0-82 13-44 C.P. 2 Sta. E Calculate the plan length, bearing and average gradient of the line AE. (L.U. Ans. 705-1 ft; 145° 11'; 1 in 22-75) 39. The following are the notes of an underground theodolite traverse. Line Azimuth Distance (ft) Vertical Angle AB 180° 00' BC 119° 01' 181-6 +15° 25' CD 160° 35' 312-0 +12° 45' DE 207° 38' 320-0 -19° 30' EF 333° 26' 200-0 -14° 12' It is proposed to drive a cross-measures drift dipping from station B at a gradient of 1 in 10 on the line of AB produced to intersect at a point X, a level cross-measures drift to be driven from station F. Calculate the azimuth and length of the proposed drift FX. (Ans. 340° 34'; 83-1 ft) 40. The following are the notes of an underground theodolite traverse: Line Azimuth Distance (ft) Vertical Angle 350 190 600 155 +28° 800 -12° It is proposed to drive a cross-measures drift to connect stations A and F. Calculate the gradient and length of the cross-measures drift, and the azimuth of the line FA. (M.Q.B./M Ans. 1 in 14-8 (3° 52'); 1391-3 ft (incl); 182°17') Bibliography MIDDLETON,R.E. and chadwick, O., A Treatise on Surveying (Spon) JAMESON, A.H., Advanced Surveying (Pitman) CLARK, D., Plane and Geodetic Surveying (Constable) RICHARDSON, P., Project Surveying (North Holland Publishing Co.) HOLLAND, J. L. WARDELL, K and WEBSTER, A.G., Surveying, Coal Mining Series (Virtue) SMIRNOFF, M. v., Measurements for Engineering and Other Surveys (Prentice-Hall) RAINSFORD, HUME F., Survey Adjustments and Least Squares (Constable) AB 089° 54' BC 150° 12' CD 180° 00' DE 140° 18' EF 228° 36' 7 TACHEOMETRY The word tacheometry is derived from the Greek tocyus swift, ^eTpov a measure. This form of surveying is usually confined to the optical measurement of distance. In all forms of tacheometry there are two alternatives: (a) A fixed angle with a variable length observed. (b) A variable angle with a fixed length observed. In each case the standard instrument is the theodolite, modified to suit the conditions. The alternatives are classified as: (1) Fixed angle: (a) Stadia systems, (b) Optical wedge systems. (2) Variable angle: (a) Tangential system - vertical staff, (b) Sub- tense system — horizontal staff. There are two forms of stadia: (1) Fixed stadia, found in all theodolites and levels. (2) Variable stadia, used in special tacheometers. 7.1 Stadia systems — Fixed stadia The stadia lines are fine lines cut on glass diaphragms placed close to the eyepiece of the telescope, Fig. 7.1. Stadia lines From Chapter 4, the basic form- ulae are: D = ms + K (Eq. 4.29) = 1 s + (f+d) (Eq. 4.28) Fig. 7.1. Diaphragm / where m = - = the multiplying constant, i f = the focal length of the object lens, i = the spacing of the stadia lines on the diaphragm, d = the distance from the object lens to the vertical axis. 359 360 SURVEYING PROBLEMS AND SOLUTIONS 7.2 Determination of the Tacheometric Constants m and K Two methods are available: (a) by physical measurement of the instrument itself, (b) by reference to linear base lines. 7.21 By physical measurement of the instrument From the general equation, D = ms + K f where m =— and K = f+d. i In the equation — = — + — (Eq. 4.19) f u v where u = the distance from the objective to the staff is very large compared with / and v and thus 1/u is negligible compared with 1/v and 1//, 1 1 / v i.e. / ^ v i.e. fn the length from the objective to the diaphragm with the focus at oo. With the external focussing telescope, this distance can be chang- ed to correspond to the value of u in one of two ways: (1) by moving the objective forward, (2) by moving the eyepiece backwards. In the former case the value of K varies with u, whilst the latter gives a constant value. The physical value i cannot easily be measured, so that a linear value D is required for the substitution of the value of / to give the factors i and K, i.e. D = sl+U+ d) i Thus a vertical staff is observed at a distance D, the readings on the staff giving the value of s. Example 7.1. A vertical staff is observed with a horizontal external focussing telescope at a distance of 366 ft 3 in. Measurements of the telescope are recorded as: Objective to diaphragm 9 in. Objective to vertical axis 6 in. TACHEOMETRY If the readings taken to the staff were 3-52, 5*35 and 7*17 ft, calculate (a) the distance apart of the stadia lines (i), (b) the multiplying constant (m), (c) the additive constant (K). From Eqs. (4.28) and (4.29), D = ms + K = l. s + (f+d) i 361 i = fs in. D-(f + d) 9-0 x (7-17-3-52) 366-25 -(0-75+0-50) = 9-0 x 3-65 366-25 - 1-25 = 0-09 in. .-. m = 1 = — = 100 i 0-09 K = f+d = 9 in. +6in. = 1-25 ft 7.22 By field measurement The more usual approach is to set out on a level site a base line of say 400 ft with pegs at 100 ft intervals. The instrument is then set up at one end of the line and stadia readings are taken successively on to a staff held vertically at the pegs. By substitution into the formula for selected pairs of observations, the solution of simultaneous equations will give the factors m and K. i.e. D, = ms, + K D„ = m s„ + K Example 7.2 The following readings were taken with a vernier theod- olite on to a vertical staff: Stadia Readings Vertical Angle Horizontal Distance 2-613 3-359 4-106 0° 150 ft 6-146 7-150 8-154 5°00' 200 ft Calculate the tacheometric constants. 362 SURVEYING PROBLEMS AND SOLUTIONS D = m (4-106 - 2-613) + K = 150 = l*493m +K =150 D 2 = m (8-154 - 6-146) cos 2 5°+ Kcos 5° = 200 = 2-008 mx 0-99620* + 0-996 20 K = 200 = 1-992 76 m + 0-99620 £ = 200 Solving these two equations simultaneously gives m = 100-05 (say 100) K = 0-7 ft N.B. The three readings at each staff station should produce a check, i.e. Middle-Upper = Lower-Middle 3-359 - 2-613 = 0-746 4-106 - 3-359 = 0-747 7-150-6-146 = 1-004 8*154-7-150 = 1-004 7.3 Inclined Sights The staff may be held (a) normal to the line of sight or (b) vertical. 7.31 Staff normal to the line of sight (Fig. 7.2) Fig. 7.2 Inclined sights with staff normal As before, D = ms + K but H = Dcos0+ SB, = Dcos0+ BE sin Q i.e. H = (ms + K)cos6+ BE sin (9 (7.1) (7.2) N.B. BE = h z = staff reading of middle line of diaphragm. BB, is - ve when 6 is a depression. Vertical difference V = Dsin0 (7.3) TACHEOMETRY 363 i.e. V = (ms + K) sin0 As the factor K may be neglected generally, H = ms cos 6 + BE sin V = ms sin (7.4) (7.5) (7.6) If the height of the instrument to the trunnion axis is h^ and the middle staff reading ft 2 ,then the difference in elevation = h,±V-h z cosd (7.7) Setting the staff normal to the line of sight is not easy in practice and it is more common to use the vertical staff. 7.32 Staff vertical (Fig. 7.3) S*Ay\ A B V 0^ ^ ^^ V ^Ke 10 HBr -<3?r H / r />i J *(ff% As before i.e. but thus Fig. 7.3 Inclined sights with staff vertical D = ms, + K = mC4,C,) + K A C are the staff readings D = m(AC cos0) + K (assuming BA^A = BC,C = 90) /k Also Also = ms cos + K (7.8) H = DcosS = ms cos z 8 + K cos (7.9) V = Dsintf = ms sin cos + K sin (7.10) V = H tan (9 (7.11) It can be readily seen that the constant K = simplifies the equations. Therefore the equations are generally modified to 364 SURVEYING PROBLEMS AND SOLUTIONS H = ms cos 2 6 (7.12) V = V 2 ms sin 26 (7.13) If it is felt that the additive factor is required, then the following approximations are justified: H = (ms +k) cos 2 6 (7.14) V = y 2 (ms+K) sin 26 (7.15) The difference in elevation now becomes = K ±V-h 2 (7.16) Example 7.3 A line of third order levelling is run by theodolite, us- ing tacheometric methods with a staff held vertically. The usual three staff readings, of centre and both stadia hairs, are recorded together with the vertical angle (V.A.) A second value of height difference is found by altering the telescope elevation and recording the new read- ings by the vertical circle and centre hair only. The two values of the height differences are then meaned. Com- pute the difference in height between the points A and B from the following data: The stadia constants are multiplying constant = 100. additive constant = 0. Remarks Staff (all measurements in ft) Point A Point B Backsights Foresig hts V.A. Staff V.A. Staff + 0°02' 6-20 4-65 3-10 + 0°20' 6-26 -0°18' 10-20 6-60 3-00 0°00' 10-37 (Aide memoir e: Height difference between the two ends of the theod- olite ray = 100s sin0 cos0, where s = stadia intercept and 6 = V.A.) (R.I.C.S.) V = 100 s sin 6 cos 6 = 50 s sin 2 6 To A, V = 50(6-20 - 3*10) sin 0°04' = +0-18 ft TACHEOMETRY 365 Difference in level from instrument axis +0*18 -4-65 -4-47 Check reading V = 50(3-10) sin0°40' = +1-80 Difference in level from instrument axis + 1-80 -6-26 -4-46 mean -4-465 B, V = 50(10-20 -3-00) sin 0°36' = -3-76 Difference in level from instrument axis - 3-76 - 6-60 - 10-36 Check level -10*37 mean -10*365 Difference in level A - B - 10-365 - 4-465 - 5-900 ft Example 7.4 The readings below were obtained from an instrument station B using an anallatic tacheometer having the following con- stants: focal length of the object glass 8 in., focal length of the an- allatic lens 4*5 in., distance between object glass and anallatic lens 7 in., spacing of outer cross hairs 0*0655 in. Instrument Height of To Bearing Vertical Stadia Readings Remarks at Instrument Angle B 4*93ft A 69°30' +5° 2-16/3-46/ 4-76 Staff held C 159°30' 0° 7-32/9-34/11-36 vertical for both observa- tions Boreholes were sunk at A,B and C to expose a plane bed of rock, the ground surface being respectively 39-10, 33-68 and 18-45 ft above' the rock plane. Given that the reduced level of B was 120*02 ft., de- termine the line of steepest rock slope relative to the direction AB. (L.U.) 366 SURVEYING PROBLEMS AND SOLUTIONS By Eq. (4.36), Then the multiplying factor m = / = 8in /, - 4-5 in x = 7 in i = 0*0655 in //, 8x4*5 0-065 5(8 + 4-5-7) = 99-93 (say 100) At station B: To A, H = 100 x 2-60 cos 2 5°= 258*02ft V = 258-02 tan 5° = +22-57 ft .'. Level of A = 120*02 + 22*57 + 4*93 - 3*46 = 144*06 ft To C, H = 100 x 4*04 = 404*00 ft V =0 .'. Level of C = 120*02 + + 4*93 - 9'34 = 115-61 ft Dip Surface level 120-02 Depth 33-68 Bed level 86-34 Surface level 144-06 Depth 39-10 Bed level 104-96 Surface level 115-61 Depth 18-45 Bed level 97-16 Fig. 7.4 Gradient AB is (104*96 - 86*34) in 258*02 ft 18*62 in 258*02 ft At point X in Fig. 7.4, i.e. on line AB where the bed level is that of C, Difference in level AC = 104*96 - 97*16 = 7*80 258*02 Length AX = 7*80 x 18*62 = 108 *09 ft TACHEOMETRY 367 BX = 258-02 - 108*09 = 149-93 ft Angle B = 159° 30' - 69°30' = 90°00' In triangle BXC, Angle BCX (a) = tan" 1 BX/BC = tan" 1 149-93/404-0 = 20° 22' Therefore the bearing of full dip is perpendicular to the level line CX, i.e. = Bearing AB + a = 69°30'+ 180°+20°22' = 269°52' 7.4 The Effect of Errors in Stadia Tacheometry 7.41 Staff tilted from the normal (Fig. 7.5) Fig. 7.5 Staff tilted from the normal If the angle of tilt B is small then 4,C, ~ AC = s 4 t C, = A 2 C 2 cosB i.e. s = s, cosjS Thus the ratio of error e = — ! — ; — = l-cos/8 (7.17) Thus the error e is independent of the inclination 0. 7.42 Error in the angle of elevation with the staff normal H = D cos + BE sin (9 N 368 SURVEYING PROBLEMS AND SOLUTIONS Differentiating gives — = - D sin d + BE cos Q 86 8H = (-D sintf + BE cos 0) Si 7.43 Staff tilted from the vertical (Fig. 7.6) (7.18) Fig. 7.6 Staff tilted from the vertical Consider the staff readings on the vertical staff at A, B and C, Fig. 7.6. If the staff is inclined at an angle /S away from the observ- er, the position of the staff normal to the line of collimation will be at XY when vertical and X^ when normal to the collimation at the inter- section with the inclined staff. Assuming that BXX, = BYC = B,X,A, = B,Y,Y ^ 90° then, with angle 6 an elevation, XY = AC cos d = s cos Q X,y, = 4,C, cos(6>+/3) = s^cos(d + fi) Assuming that XY ~ X, Y, , then s cos — s, cos (0 + /S ) s, cos (0 + /3) s = COS0 (7.19) i.e. the reading s on the staff if it had been held vertically compared with the actual reading s, taken on to the inclined staff. Similarly; if the staff is inclined towards the observer, s, cos (0 - /S) s = COS0 (7.20) TACHEOMETRY 369 If the angle 6 is a depression the equations have the opposite sense, i.e. s, cos (0 - B) Away from the observer s = (7.21) COS0 ^ , , s i cos (.0 + fi) Towards the observer s = ' — (7.22) cos 8 Thus the general expression may be written as s, cos (0 ±B) s = J 1—Ti (7.23) cos 6 The error e in the horizontal length due to reading s, instead of s is thus shown as True length = H T = ms cos 2 6 ms, cos (d ± 8) cos 6 Apparent length = H A = ms, cos 2 cos 2 (7.24) Tcos (6 ±B) 1 Error e = H T -H A = ms. cos 2 (9 ^^ ~ - 1 (7.25) L cos 6 J Hf — H A The error expressed as a ratio = ms. cos 2 d cos (6 ±8) cos 6 ms^ cos 2 cos(0 ±B) cos - 1 (7.26) cos 6 cos yS + sin 6 sin /3 - cos 6 cos 6 = cos B ± tan sin /3 - 1 If /3 is small, <5°, then e = B tantf. Example 7.5 In a tacheometric survey an intercept of 2*47 ft. was recorded on a staff which was believed to be vertical and the vertical angle measured on the theodolite was 15°. Actually the staff which was 12 ft long was 5 in out of plumb and leaning backwards away from the instrument position. Assuming it was an anallatic instrument with a multiplying con- stant of 100, what would have been the error in the computed horizon- tal distance? 370 SURVEYING PROBLEMS AND SOLUTIONS In what conditions will the effect of not holding the staff vertical but at the same time assuming it to be vertical be most serious? What alternative procedure can be adopted in such conditions ? By Eq. (7.19), Thus s = Fig. 7.7 s, cos (6 + j8) COS0 /8 = tan -1 5712 x 12 = 1°59'20" s = 2-47 cos (15° + 1° 59' 20") cos 15° = 2-4456 ByEq.(7.12), H = ms cos 6 8H = m cos 2 8s = 100 x cos 2 15° x (2-47 - 2'4456) = 2-44 cos 2 15 = 2-28 ft Alternatively, By Eq.(7.25), 8H = ms a J cos(g + /B ) , cos 2 1 247 cos 2 -[ COS0 cos 16°59'20' cos 15° = 230-130 3 [0-99009-1] = 2-28 ft TACHEOMETRY 371 7.44 Accuracy of the vertical angle 6 to conform to the overall accuracy (Assuming an accuracy of 1/1000) From H = ms cos 2 6 differentiation gives 8H = -2mscos 6 sin 6 86 3H 1 2ms cos d sin d 86 For the ratio H 1000 ms cos 2 6 cos 6 86 = 2 sin0 x 1000 1 cot 6 2000 Tf „ „ rtQ 206 265 cot 30° If (9=30°, 86 = — seconds = 178 seconds; i.e. ~ 3 minutes N.B. 1 in 1000 represents 0*1 in 100 ft. The staff is graduated to 0*01 ft but as the multiplying factor is usually 100 this would repres- ent 1ft. If estimating to the nearest 0*01 ft the maximum error = ±0*005 ft. Thus taking the average error as ±0*0025 for sighting the two stadii, Average error = 0*0025 V 2 = ±0*0035 Error in distance (H) due to reading = ±0*0035 m, cos 2 6 The effect is greater as 6 — * Thus, if m = 100, 8H = ±0*35 ft If H = 100 ft 8H ^ 1 H 300 From V = D sin 6 8 V = D cos 6 86 8V Dcos6 86 ~V = D sin 6 = cot0 86 8V 1 If 77= 77^^= c ^6 86: V 1000 when = 45°, 206 265 86 = ' = 206 sec = 3min 26 sec 372 SURVEYING PROBLEMS AND SOLUTIONS when 6 = 10°, 3d 206 265 tan 10° 206 265 x 0-176 3 1000 1000 = 36 sec 7.45 The effect of the stadia intercept assumption (i.e. assuming BA^A = B&C = 90°, Fig. 7.8) 90-(8+oc) 90- Oi Fig. 7.8 Let the multiplying factor m = 100 Then a = tan '- — 206265 sec 200 200 = 0°17'11'35" la = 0°34'23" In triangle BA % A A^B = In triangle BC^ C BC, = s , sin [90 - (6 + a) ] sin (90 + a) s , cos (d + a) s, (cos d cos a - sin 6 sin a ) cos a cos a s 2 sin[9O-(0-a)] sin (90 -a) s 2 cos(0-a) s 2 (cos# cos at + sin sin a) cos a cos a s, (cos 6 cos a - sin 6 sin a ) s,(cos 6 cos a + sin sin a ) 1 T cos a cos a = s (cos - sin tan a) + s 2 (cos + sin 6 tan a ) i4 1 C 1 = (s, + s 2 ) cos + (s 2 - s,)(sin tana) (7.27) TACHEOMETRY 373 Thus the accuracy of assuming A X C^ = AC cos 6 depends on the sec- ond term (s 2 - s,) (sin 6 tana). Example 7.6 (see Fig. 7.8) If 6 = 30°, FB = 1000 ft., m = 100 and K = 0, ^ B = BC < = loo = 5 ' 0ft s. = i4,B cos - sin tan a 5-0 0-866 - 0-5 x 0*005 5-0 0-866 - 0-0025 5-790 4 Similarly BC, cos 6 + sin tan a 5-0 0-866 + 0-0025 = 5-757 1 Therefore the effect of ignoring the second term (s 2 -s;) (sin tana) = (5-790 4 - 5-757 1) (0-002 5) - -0-033 3x0-0025 = -8-325 x 10~ 5 The inaccuracy in the measurement FB thus = -8-325 x 10~ 2 ~ 0-1 ft in 1000 ft and the effect is negligible. Thus the relative accuracy is very dependent on the ability to es- timate the stadia readings. For very short distances the staff must be read to 0*001 ft, whilst as the distances increase beyond clear read- ing distance the accuracy will again diminish. Example 7.7 A theodolite has a tacheometric multiplying constant of 100 and an additive constant of zero. The centre reading on a vertical staff held at a point B was 7*64 ft when sighted from A. If the vertical 374 SURVEYING PROBLEMS AND SOLUTIONS angle was +25° and the horizontal distance A B 634*42 ft, calculate the other staff readings and show that the two intercept intervals are not equal. Using these values, calculate the level of B if A is 126*50 ft A.O.D. and the height of the instrument 4*50 ft. (L.U.) 90-(B + oe) 90+<* 90-otT Horizontal distance Inclined distance mcos 2 d 100 cos 2 25° = 7-72 ft = HD sec 6 = 634-42 sec 25° = 700-00 ft = 3-50 s n cosa cos (0 + a) 3-50 cos 0°17'11" cos 25° 17 '11" = 3-87 (sine rule) TACHEOMETRY 375 Similarly, S 2 s cos a cos(0-a) 3-50 cos 0°17'11" cos 24°42'49" = 3^85 Check 3-87 + 3-85 = 7-72 ft Staff readings are 7-64 7-64 + 3-87 and -3-85 Upper 11-51 Lower 3-79 -3-79 Check s 7-72 Vertical difference = HDtand = 634-42 tan 25° = +295-84 ft Level of A = 126-50 + 295-84 + 4-50 + 426-84 - 7-64 Level of B = +419-20 (sine rule) Example 7.8 Two sets of tacheometric readings were taken from an instrument station A, the reduced level of which was 15-05 ft., to a staff station B. (a) Instrument P — multiplying constant 100, additive constant 14-4 in, staff held vertical. (b) Instrument Q — multiplying constant 95, additive constant 15-0 in, staff held normal to line of sight. Inst At To Height of Inst. Vertical Angle Stadia Readings P A B 4-52 30° 2-37/3-31/4-27 Q A B 4-47 30° What should be the stadia readings with instrument Q? (L.U.) To find level of B (using instrument P) By Eq. (7.10), V = ms sin B cos 6 + K sin 6 376 SURVEYING PROBLEMS AND SOLUTIONS V, = 100 x (4-27 - 2-37) sin 30° cos 30° +1-2 sin 30 = 190 x 0-5x0-8660+ 1-2x0-5 = 82-27 + 0-60 = 82-87 ft By Eq. (7.9), tf, = ms cos 2 + K cos = 190 x 0-866 03 2 + 1-2 x 0-866 03: = 142-49+ 1-03 - 143-52 ft Also by Eq. (7.11), V,= // 1 tan0 = 143-52x0-57735 = 82-86 ft (Check) Level of B = 15-05 + Ht of inst + V - middle staff reading = 15-05 +4-52+ 82-87 -3-31 = 99-13 ft Using instrument Q In Fig 7. 2, V = (H - BE sin 6) tan 6 V 2 = (143-52 - BE sin 30°) tan 30° = 143-52 x 0-577 35 - BE x 0-5 x 0-577 35 = 82-86 - 0-28868 BE Level of B = 15-05 + 4-47 + V 2 - BE cos = 99-13 = (82-86 -0-288 68 BE) -0-866 03 BE = 79-61 -1-15471 BE =-3-25 BE = -3-25 1-154 71 middle reading =■■ 2-81 By Eq. (7.5), H 2 = ms cos 6 + BE sin d = 95 x 0-866 03 s+ 2-81x0-5 = 143-52 = 82-27 s = 143-52- 1-40 142-12 s = = 1-727 82-27 l / 2 s = 0-86 '.Readings are 2-81 ±0-86 = 1-95/2-81/3-67 Example 7.9 Three points A,B and C lie on the centre line of an existing mine roadway. A theodolite is set up at 6 and the following observations were taken on to a vertical staff. TACHEOMETRY 377 Staff at Horizontal Vertical Staff Readings Circle Circle Stadia Collimation A 002°10'20" +2°10' 6-83/4-43 5-63 C 135°24'40" -1°24' 7*46/4-12 5-79 If the multiplying constant is 100 and the additive constant zero calculate: (a) the radius of the circular curve which will pass through A,B and C. (b) the gradient of the track laid from A to C if the instrument height is 5-16. (R.I.C.S.) Fig. 7.10 Assumed bearing BA = 002°10'20" BC = 135°24'40" Angle ABC = 133° 14' 20" Angle AOC = 360 - 2(133° 14 '20") = 93°31 , 20" Line AB Horizontal length (H) = ms cos Q = 100(6-83-4-43) cos 2 2° 10' = 240 cos 2 2° 10' - 239-66 ft 378 SURVEYING PROBLEMS AND SOLUTIONS Vertical difference (V) = H tand = 239-66 tan 2° 10' = +9-07 ft Line BC H = 100(7-46 -4-12) cos 2 1°24' = 333-80 ft V = 333-80 tan 1° 24' = 8-16 ft In triangle ABC A-C a- c A + C Tan — -— = tan a + c 2 333-80 - 239-66 180 - 133° 14 '20" 333-80 + 239-66 94-14 tan 573-46 A - C = 4°03'35" 2 A -i- C ^—^ = 23°22'50" 2 A = 27° 26 '25" C = 19°19'15" AB tan 23°22'50' = 2R (sine rule) sin C 239-66 R = Differences in level 2 sin 19° 19 '15" = 362-18 ft BA = 5-16 + 9-07-5-63 = +8-60 BC = 5-16-8-16-5-79 = -8-79 AC = 17-39 Length of arc AC = 362-18 x 93°31' 20" rad = 362-18 x 1-632 271 = 591-18 ft TACHEOMETRY 379 Gradient = 17-39 ft in 591-18 ft = 1 in 34 Example 7.10 The following observations were taken during a tacheo- metric survey using the stadia lines of a theodolite (multiplying const- ant 100, no additive constant.) Station Set Station Staff Readings Vertical Bearing at Observed U M L Angle B A 5-62 6-92 8-22 +5°32' 026°36' C 3-14 4-45 5-76 -6°46' 174°18' Calculate (a) the horizontal lengths AB and BC. (b) the difference in level between A and C. (c) the horizontal length AC. Line BA s = 8-22-5-62 = 2-60 Horizontal length = 100 s cos 2 6 i 1 */ \ = 100 x 2-60 cos 2 5< 257-58 ft D 32' L $/ i y 1 Vertical difference = HtanO 257-58 tan 5°32' + 24-95 ft if \ Line BC s = 5-76-3-14 = 2-62 8< ( Jl47 # 42' / H = 100x2-62 cos 2 6* 258-36 ft >46' \ «■* i V = 258-36 tan 6°46' -30-66 ft \ -* > \* 1 \ ^* ' \ jo | Native levels A A + 24-95 - 6-92 + 18-03 above B \<p / \& i -30-66 - 4-45 c C -35-11 below B Fig. 7.11 Difference i n level A-C 53' 14 ft 380 SURVEYING PROBLEMS AND SOLUTIONS In triangle ABC, A-C 258-36 - 257-58 180 - 147°42 / Tan 2 = 258-36 + 257-58 ta " 2 A-C 2 A + C 0°01' 16°09' A = 16° 10' Length AC = 258-36 sinl47°42' cosec 16°10' = 4 95-82 Exercises 7(a) 1. P and Q are two points on opposite banks of a river about 100 yd wide. A level with an anallatic telescope and a constant of 100 is set up at A on the line QP produced, then at B on the line PQ produced and the following readings taken on to a graduated staff held vertically at P and Q. What is the true difference in level between P and Q and what is the collimation error of the level expressed in seconds of arc, there being 206 265 seconds in a radian. To Staff readings in feet From Upper Stadia Collimation Lower Stadia A P Q 5-14 3-27 4-67 1-21 4-20 below ground B P Q 10-63 5-26 8-51 4-73 6-39 4-20 (I.C.E. Ans. 3-62 ft; 104" above horizontal) 2. Readings taken with a tacheometer that has a multiplying const- ant of 100 and an additive constant of 2*0 ft were recorded as follows: Instrument at Staff at Vertical Angle Stadia Readings Remarks Q 30°00' elevation 5-73, 6-65, 7-57 Vertical staff Although the calculations were made on the assumption that the staff was vertical, it was in fact made at right angles to the collimation. Compute the errors, caused by the mistake, in the calculation of horizontal and vertical distances from the instrument to the foot of the staff. Give the sign of each error. If the collimation is not horizontal, is it preferable to have the staff vertical or at right angles to the collimation? Give reasons for TACHEOMETRY 381 your preference. (I.C.E. Ans. Horizontal error -24*7 ft, Vertical error -13'2 ft) 3. The following readings were taken with an anallatic tacheometer set up at each station in turn and a staff held vertically on the forward station, the forward station from D being A. Station Height of Instrument Stadia Readings Inclination (elevation + ve) A B C D 4-43 4-61 4-74 4-59 4-93 3-54 2-15 5-96 4-75 3-54 5-15 3-72 2-29 6-07 4-64 3-21 + 0°54' -2°54' + 2°48' -1°48' The reduced level of A is 172-0 ft and the constant of the tacheo- meter is 100. Determine the reduced levels of B, C and D, adjusted to close on A, indicating and justifying your method of adjustment. (I.C.E. Ans. 177-5; 165-4; 180-7) 4. The focal lengths of the object glass and anallatic lens are 5 in and 4y 2 in respectively. The stadia interval was 0-lin. A field test with vertical staffing yielded the following: Instrument Staff Staff Vertical Measured Horizontal Station Station Intercept Angle Distance (ft) P Q 2-30 +7°24' 224-7 R 6-11 -4°42' 602-3 Find the distance between the object glass and anallatic lens. How far and in what direction must the latter be moved so that the multiplying constant of the instrument is to be 100 exactly. (L.U. 0*02 in away from objective) 5. Sighted horizontally a tacheometer reads r, = 6'71 and r 3 = 8'71 on a vertical staff 361*25 ft away. The focal length of the object glass is 9 in. and the distance from the object glass to the trunnion axis 6 in. Calculate the stadia interval. (I.C.E. Ans. 0*05 in) 6. With a tacheometer stationed at X sights were taken on three points, A, B, and C as follows: nstrument To Vertical Stadia Remarks at Angle Readings X A -4°30' 7-93/6-94/5-95 R.L. of A = 357-09 (Staff nor- mal to line of sight B 0°00' 4-55/3-54/2-54 R.L. of B= 375-95 (Staff vertical) C + 2°30' 8-85/5-62/2-39 Staff vertical 382 SURVEYING PROBLEMS AND SOLUTIONS The telescope was of the draw-tube type and the focal length of the object glass was 10 in. For the sights to A and J5, which were of equal length, the distance of the object glass from the vertical axis was 4*65 in. Derive any formulae you use. Calculate (a) the spacing of the cross hairs in the diaphragm and (b) the reduced level of C. (L.U. Ans. 0-102 in.; 401-7 ft) 7. The following readings were taken on a vertical staff with a tach- eometer fitted with an anallatic lens and having a constant of 100: Staff Station Bearing Stadia Readings Vertical Angle A 27°30' 2-82 4*50 6*18 + 8°00' B 207°30' 2-54 6*00 9-46 -5°00' Calculate the reduced levels of the ground at A and B, and the mean slope between A and B. (L.U. Ans. +41-81; -66-08; 1 in 9*42) 8. Tacheometric readings were taken from a survey station S to a staff held vertically at two pegs A and B, and the following readings were recorded: Point Horizontal Circle Vertical Circle Stadia Readings 62°00' 152°00' + 4° 10 '30" -5°05'00" 4-10/6-17/8-24 2-89/6-17/9-45 The multiplying constant of the instrument was 100 and the addi- tive constant zero. Calculate the horizontal distance from A to 8 and the height of peg A above the axis level of the instrument. (I.CE. Ans. 770-1 ft; 23-89 ft) 9. In a tacheometric survey made with an instrument whose constants were f/i :*= 100, (f+d) = 1*5, the staff was held inclined so as to be normal to the line of sight for each reading. How is the correct inclin- ation assured in the field? Two sets of readings were as given below. Calculate the gradient between the staff stations C and D and the reduced level of each. The reduced level of station A was 125*40 ft. Instrument Staff Height of Azimuth Vertical Stadia Readings at at Instrument Angle A C 4-80 44° + 4°30' 3-00/4-25/5-50 D 97° -4°00' (L. 3-00/4-97/6-94 U. Ans. 1 in 6-57) TACHEOMETRY 383 10. (a) A telescope with tacheometric constants m and c is set up at A and sighted on a staff held vertically at B. Assuming the usual relationship D = ms + c derive expressions for the horizontal and vertical distances between A and B. (b) An instrument at A, sighted on to a vertical staff held at B and C, in turn gave the following readings: Sight Horizontal Circle 05°20' 95°20' Vertical Circle + 4°29'00" -0°11'40" Staff Readings (ft) 1-45/2-44/3-43 2- 15/3- 15/4- 15 If the instrument constants are m = 100, c = 0, calculate the gradient of the straight line BC. (N.U. Ans. 1 in 16-63) 7.5 Subtense systems 7.51 Tangential method (with fixed intercept s and variable vertical angles a and /S) From Fig. 7.12 Fig. 7.12 Tangential method DC = y = //tana AD = s + y = H tan /8 AC = s = H (tan - tana) s H = Alternatively, as tan /3 - tan a 7 = /3-a, s ///cos a (7.28) sin 7 sin(90-j8) H = s cos a cos /8 cosec y (7.29) 384 SURVEYING PROBLEMS AND SOLUTIONS This equation (7.29) was modified by M. Geisler (Survey Review, Oct. 1964) as follows: s H = cos a cos (y + a) siny s cos (2 a + y) + cos y siny siny 1 + cos y) - { 1 - cos (2 a + y) } ay • 2 cos -i- - sin 2 y y 2 sin— cos— L 2 2 = -s cot— - —s cot- ' 2 2 2 2 y As y is small, cos 2 -^ ^ 1. y Also a+-y = i y l y 2 H = -~s cot« - ys cot-j sin i y = -^scot^- (l-sin 2 0) 1 y = — s cot— cos 2 2 2 (7.30) Alternatively, the above equation may be derived by reference to Fig. 7.13. 1 y H x= ^ s i cot ~2~ where s i = ^i C i s, ~ s cos 6 (assuming A,AB and BC X C are similar figures) H = H.cosd 1 y H = —s cot— cos 2 6 as above but and 1 y N.B. In the term -^s cot-~ .2/ 7^ sin (a + ~2 J 2 y COS 4r cot -^ is very large, TACHEOMETRY 385 Fig. 7.13 so that any approximation to cos 2 Z is greatly magnified and the following approximation is preferred: As before H = —r—r, cos a cos ( y + a) siny s ■cos K)~K) (—-?) but cosy ~ 1 sin y \ II \ 2 s f cos 20 + cosy siny [ r 2 i ~i H H [" cos 20+1 1 siny s cos 2 ^ siny s cosecycos 2 (7.31) Geisler suggests that by using special targets on the staff, thus ensuring the accuracy of the value of s, and the use of a 1" theodol- ite, a relative accuracy up to 1/5000 may be attained. He improved the efficiency of the operation by using prepared tables and graphs rela- tive to his equation. The accuracy of the method is affected by: (1) An error in the length of the intercept s. (2) An error in the vertical angle. (3) Tilt of the staff from the vertical. 386 SURVEYING PROBLEMS AND SOLUTIONS (1) Error in the intercept s This depends on (a) error in the graduation, (b) the degree of precision of the target attachment. 8H = ^-^ (7.32) s (2) Error in the vertical angle From Eq. (7.28), H ^ tan jS - tan a + s sec a 8a (tan /3 - tan a ) -s sec 2 /8 8B 8H„ = , — ^— £- (7.33) P (tanjS-tana) 2 total r.m.s. error 8H (tan/3 - tana) = y/8H* + S/J 2 = Jse^a&^ + sec^S/S 2 ] 1 " /'fan R _ tan^ 2 If 5a = 83, s 5a ^ = -r. — a — : r; [sec 4 a + sec 4 B ] (tan B - tan a) 2 ^ -^f.[sec 4 a + sec* BY (7.34) If a and B are small, yj2H z 8a 8H = V s (7.35) (3) Tilt of the staff The effect here is the same as that described in Section 7.43, i.e. s,cos(0 ±B) s = cos where B = tilt of the staff from the vertical. Example 7.11 A theodolite was set over station A, with a reduced level of 148-73 ft A.O.D., the instrument height being 4-74ft. Obser- vations were taken to the 10 ft and 2 ft marks on a staff held vertical at three stations with the following results: Instrument Station Station Observed Vertical Angles Top Bottom A B +9°10' +3°30' A C +1°54' -2°24' A D -5°15' -12°10' TACHEOMETRY 387 Find the distance from A to each station and also their reduced levels. (E.M.E.U.) By Eq. (7.28), Horizontal distance AB = tan /8 - tan a 8 Vertical distance Level of B Horizontal distance AC = Vertical distance Level of C tan9°10' -tan3°30' 8 0-100 21 = Afitan3°30' = 4-883 ft = 4-88 + 4-74+ 148-73 -2-00 ft = 156-35 ft A.O.D. 8 tanl°54' +tan2°24' - 106-5=1 ft 0-07509 = -106-55 tan 2° 24' = -4-47 ft = 148-73 + 4-74 - 4-47 - 2-00 147-00 ft A.O.D. 8 Horizontal distance AD = tanl2°10'-tan5 o 15' 8 0-12371 = 64-53 ft Vertical distance = -64-53 tan 12°10' = -13-91 ft Level of D = 148-73 + 4-74 - 13-91 - 2-00 ft = 137-56 ft A.O.D. Alternative solutions ByEq.(7.29), H 8cos9°10' cos3°30' Horizontal distance AD = — : — ,„.,-, ^^^ sin (9° 10 - 3°30 ) 8 x 0-987 23 x 0-998 14 0-098 74 = 79-84 ft or by Eq. (7.30), AB = % x 8 x cot 2°50' x cos 2 V 2 (12°40' ) = 4cot2°50' cos 2 6 o 20' 388 SURVEYING PROBLEMS AND SOLUTIONS = 4x20-2056x0-99390 = 79-84 ft 7.52 Horizontal subtense bar system (Fig. 7.14) Theodolite Horizontal angle (oc) measured by the theodolite Fig. 7.14 Horizontal subtense bar system The horizontal bar of known length b, usually 2 metres, is set perpendicular to the line of sight TB. Targets at A and C are successively sighted and the angle a, which is measured in the horizontal plane, recorded. The horizontal distance TB, = H is then obtained b a H = - cot - 2 2 If the bar is 2 metres long, H = cot— metres (7.36) (7.37) The horizontal angle a is not dependent on the altitude of the bar relative to the theodolite. N.B. As the bar is horizontal, readings on one face only are necess- ary. Factors affecting the accuracy of the result are (1) The effect of an error in the subtended angle a By Eq.(7.36), b a H = j cot - TACHEOMETRY 389 b 2 tan — a a If a is small, then tan——— radians. a Differentiating with respect to a, b H = - (7.38) SOa - ~ (7-39) ft but a=- 8Ha = -uya (7 40) b lhe error ratio -rr- = n a where 8a and a are expressed in the same units. If 8a = +1" and b = 2m, then by Eq. (7.40), 8// = ± metres 2 x 206 265 (7.41) 8H = ± 412^0 metreS (742) * Slf^oo metres (743) Example 7.12 To what accuracy should the subtense angle a be measured to a bar 2 metres long if the length of sight is approximately 50 metres and a fractional error of 1/10 000 must not be exceeded? By Eq. (7.41), SH 8a 1 but H a 10000 a 5a = 10000 b a =ll 8a 10000// 390 SURVEYING PROBLEMS AND SOLUTIONS 2 x 206 265 = ^OOOOxSO 8600 ^ 5 = + 0-825 " Example 7. 13 If the measured angle a is approximately 2°, how accurately must it be recorded if the fractional linear error must not exceed 1/10 000? 8H 8a 1 H a 10000 8a = 2 x 3600 10 000 = ±0-72" an error in th e length o H = b a 8H b = 8b a H = 8b b (2) The effect From Eq.(7.38), (7.44) (7.45) If the error ratio is not to exceed 1/10000, then ,?3 = §k 1 H b 10000 As the bar = 2 metres = 2000 mm, 8b is limited to 0-2 mm. The bar is usually made of invar steel and guaranteed by the manu- facturer to ±0-05 mm. It requires a change of 20°C for these limits and thus for most uses the bar is considered constant. (3) The effect of an error in the orientation of the bar Consider half the bar, Fig. 7. 15. Fig. 7. 15 Orientation of the bar TACHEOMETRY 391 Let the half bar AB be rotated through an angle 6 to A : B. The line of sight will thus be assumed to be at A z . r, b . e u . AAi = 2x — sin- = osm— AA 2 = 8H AA y sin M sin T osin-^jsin ycos^- 6 . j8 + cos«- sin-|j- sin -£- 2 = b sin z x-cot— + b sin-^ cos ■ 2 2 2 2 P but bcot±-~2f/ 2 Q J. 8H = 2H sin 2 - +- sin0 (7.46) Neglecting the second term as both —and 6 are small, -f =2sin 2 -^ (7.47) If the relative accuracy is limited to 1/10000, then m 1 2sin 2 -2- H 10000 2 sin 2 = 5 x 10~ 4 $ = 1°17' As the bar usually has a sighting device, it can be oriented far more accurately than to the above limit, and this non-rigorous analysis shows that this effect can be ignored. The accuracy of the whole system is thus entirely dependent on the angle a . Assuming the angle a can be measured to ±1" and the bar is 2 metres, da 1 a 10000 a = 10000' 392 SURVEYING PROBLEMS AND SOLUTIONS H = 2 x 206 265 a " 10000 = 41-25m metres For most practical purposes, for an accuracy of ±lcm, the dis- tance can be increased to 75 metres. To increase the range of the instrument various processes may be used, and these are described in the next section. 7.6 Methods used in the.field 7.61 Serial measurement (Fig. 7.16) Fig. 7.16 Serial measurement tj 2 = //, +h 2 + ... = h b = ~2l By Eq. (7.40), 8H, = Total r.m.s. error = y/2*(5H) z = [ cot T +cot f + -] H, 8a i H;8a 2 +... H If ff, =H 2 = H n = — , a, = a 2 = a n and 8a, = 8a 2 = 8 n , 2SH = ±\/n H 2 8a n z b = +■ H 2 8a n^b If b = 2 metres, 5a = ±1" and f/ = n//, Total 8H = ± 2 x 206 265 x n 3 = ± 412 530 n 3 (7.48) (7.49) (7.50) The error ratio TACHEOMETRY H Z + 400000n 3/2 8H „ H H 400000/x 372 8H If 5a = ±1", b = 2 metres, n = 2 and ~ = 1/10000 then /7 400000 2 10000 400000x2-83 H = = 113 metres 10000 393 (7.51) (7.52) 7.62 Auxiliary base measurement (Fig. 7.17) For lines in excess of 150m, an auxiliary base of 20 — 30 m may be set out at right angles to the traverse line, Fig. 7.17. Fig. 7.17 Auxiliary base measurement Angles a and /3 are measured Differentiating, % b a = — cot — 2 2 H = }% cotjS b a n = ~2 cot 2 Cot ^ = — cotjS 8H n Sty = bcotjS 5a Z?cosec 2 j3 5/3 (7.53) (7.54) 394 SURVEYING PROBLEMS AND SOLUTIONS Total r.m.s. error 8H If a and /8 are both small, 8H = b 2 cot 2 £ 5a bcosec 4 ft 8(3 2 but H = 8H = b 8a b 2 8fi z * I a A (3 2 + a 2 F J b a/3 ~H 2 8a 2 2 a + H 2 8fi 2 P 2 J (7.55) (7.56) (7.57) If a = /8 and 5a = 5j3, 8H = b H As H = - and H = -2- * a and as a = j8, \/2 # Sa and ^ = A H, = s/(bH) b 8H = y/(bH) V2f/ 3/2 Sa V H (7.58) (7.59) (7.60) If b = 2 metres and 8a = ±1", 8H = H 3 and the fractional error If __ = 1/10 000, H 206 265 H " 206 265 (7.61) (7.62) then TACHEOMETRY y/H = 20-626 5 H = 410 metres 395 the sub-base H b = \/(2H) = V(2x410) = 28*7 metres 7.63 Central auxiliary base (Fig. 7.18) For lines in excess of 400 metres, a double bay system may be adopted with the auxiliary base in the middle. Fig. 7.18 Central auxiliary base Length TJ 2 = H = H, + H z b ,ol , a b ^ a. , _ = — cot — cot p. + — cot — cot /3„ 2 2 ! 2 2 2 = ^cot^[cota + cotaj 2 2' 2 = *[cot# + cot/6 2 ] a If a , j8, and j8 2 are each small, Differentiating, (7.63) (7.64) (7.65) "*».--£#. 396 SURVEYING PROBLEMS AND SOLUTIONS b 8Hn = --^r<5)3 2 "2 a/3 2 2 Total t.m.s. error 8H = y/{8fg +8H/3* + 8Hfi If a = ft = j^ and 5a = Sft = <5/8 2 , then b /fea 2 /2 2 1 1 as\ (a \a a a V6 b8a but by Eq. (7.65) H = a = 8H = 2b a 2 lb (ff, = #2) ^/6H 3/2 b8a 2f /2 ff 2 V3 fl 3/2 Sa 2yjb If b= 2m and 5a = ±1", V3/T 2V2x 206 265 H 3/2 336 818 ,3/2 If SH/ff = 1/10000, <5tf z. 8H W 350 000 y/H 10000 350000 V^ - 35 H ~ 1225 metres The auxiliary base f^ = H b = \/H = 35 metres. (7.66) (7.67) (7.68) (7.69) (7.70) TACHEOMETRY 397 7.64 Auxiliary base perpendicularly bisected by the traverse line (Fig. 7.19) Fig. 7. 19 Auxiliary base bisected by the traverse line Here u b .a H = I cot J H = — cot — 1 2 2 H 2 = y cot T H = H + H 2 = — cot — + cot 2 L 2 2 J TCOtT 4 2 cot—- + cot — 2 2 If a , /3, and 6 2 are all small, H = 2a "2 2 " />[ i li a ^ Ii + K„ s/t = a "i i 5a (7.71) (7.72) 8H 3 = — SB. 1 aff 5/Z/9 = "ft % 398 SURVEYING PROBLEMS AND SOLUTIONS 2 cw, 2 Total r.m.s. error 8H =y/{Sf£ + 8Hp* + SH/3, MfeVl IT 8 A ^ If a = /3, = j8 2 and da = 8/3, = S&,, a . / 1 „ 2 „2 8H = da 4 8a Sa a a a a V6 b8a (7.73) but H = 2b 2b J 2b a = J~H V6 b// 3/2 5a 8tf = 2 3/z b 3/2 V3 J/ 3 ' 2 8a 2V^ (7.74) N.B. This is the same value as for the central auxiliary base (7.68). 7.65 With two auxiliary bases (Fig. 7.20) The auxiliary base f^ is extended twice to H. Here Fig. 7.20 Two auxiliary bases K = —cot — 6 2 2 //, = H b cotjS H = H x cot<£ TACHEOMETRY H = — cot— cot /8 cot H = -cot/3 cot0 If a , /3 and are all small, H = 6 a/80 SH = b§a ~a 2 /80 BHp = b5/8 a/3 2 5fk - bd</> * a/30 2 Total r.m.s. error 8H = yJSHj +8H 2 p + SHj b l\8a 5/8 2 50' a/80/s/ i a 2 /S 2 If a = /8 = and 5a = 5/8 = 50, i.e. then H b //, H b H b H, », = VftH H b ", 2 ,_ = — = JUT H v ' 8H b /35a 2 = a 3 */ a 2 V3 b8a 399 (7.75) (7.76) (7.77) but 8H a = 3 Jw V3b5aH 4/3 b 4/3 4/3, V3/^ /3 5a J/3 (7.78) 400 SURVEYING PROBLEMS AND SOLUTIONS If b= 2m and da = ±1", 8H = 8H If — = 1/10000, H and y/3H 4 206 265 x H A/3 150000 '\/2 8H = 1 *JH H 10 000 150000 3/ = 15 H = 3375 metres H b = b_ a , 1/3 b w (7.79) fi = * a/3 r ,i/3 H' = 2 2/3 H U3 = 1-5866 x 15 = 23'8 metres 7.66 The auxiliary base used in between two traverse lines (Fig. 7.21) Fig. 7.21 Auxiliary base between two traverse lines b a H b = -cot- H. = % sinCft+fl,) sin/3, TACHEOMETRY 401 b a 2 cot 7j sin (j8, + 0,) sin/3, bsinQ3 1 + 0,) (7.80) (7.81) Similarly «. = a sin /3, ft a 2 cot g sin (p 2 + #2) sin /3 2 fcsin(j3 2 +<9 2 ) a sin j8 2 Here the errors are not analysed as the lengths and angles are variable. Example 7.14 A colliery base line AB is unavoidably situated on ground where there are numerous obstructions which prevent direct measurement .It was decided to determine the length of AB by the method illustrated in Fig. 7.22, where DE is a 50 metre band hung in catenary with light targets attached at the zero and 50 metre marks. From the approximate angular values shown, determine the maxi- mum allowable error in the measurements of the angles such that the projection of error due to these measurements does not exceed: (a) 1/200000 of the actual length CD when computing CD from the length DE and the angle DCE and (b) 1/100 000 of the actual length AB when computing AB from the angles ACD, CDA, BDC, and DCB and the length DC. For this calculation, assume that the length DC is free from error. (R.I.C.S.) D 50m E 'QOSvV ^ 70^~ "^70°/ A S i3>\ r 1 0(J V. 70^_ ■fJO*^ c Fig. 7.22 402 SURVEYING PROBLEMS AND SOLUTIONS Assuming Angle DCE(a) = Angle DAB (-|)= 1 / 2 (180-2x70) = 20°, (a) DC = DE cot a (b) The error 8DC = DE 2 * cosec a 8a TU ♦• 8DC The error ratio DC = DE cosec a 8a DE cot a = 8a 1 sin a cos a 200 000 8a = 206 265 x Vi sin 2a 200000 = 0-51566 sin 40° = 0-33 seconds (say 1/3") DC \ ft ft] AB = ~J cot T+ cot-j £C 2 ft ^ 8AB R = ±— cosec y 5/8, % DC ft •546^ = ± — cosec —SB. Total error SAB 2 DC ft cosec 4 — 1 5/3f + cosec 4 — -8B% 2 but ft = ft and assuming 5ft = 5ft. The error ratio 8 A B = ± — V 2 cosec 2 — 5/8 SAB DC/2 y/2 cosec 2 B/28B ~AB = DC/2 x 2 cot 0/2 V2 5/3 = 2 sin ft/2 cos jS/2 V25/3 _ 1 = sin/6 = 100000 206 265 sin 40° 5/3 = V2x 100000 = 0*94 seconds (say ± 1") TACHEOMETRY 403 Exercises 7(b) 11. (i) What do you understand by systematic and accidental errors in linear measurement, and how do they affect the assessment of the probable error? Does the error in the measurement of a particular distance vary in proportion to the distance or to the square loot of the distance? (ii) Assume you have a subtense bar the length of which is known to be exactly 2 metres (6*562 ft) and a theodolite with which horizontal angles can be measured to within a second of arc. In measuring a length of 2000 ft., what error in distance would you get from an angular error of 1 second? (iii) With the same equipment, how would you measure the dis- tance of 2000 ft in order to achieve an accuracy of about 1/5000? (Aide memoire: 1 second of arc = 1/206265 radians.) (I.C.E. Ans. (ii) ±2-95 ft ) 12. (a) When traversing with a 2 metre subtense bar, discuss the methods which can be adopted to measure lines of varying length. In- clude comments on the relative methods of angular measurement by repetition and reiteration. (b) A bay length AB is measured with a subtense bar 2 metres in length, approximately midway between and in line with AB. The mean angle subtended at A = 1°27'00" at B = 1°35'00" Calculate the length AB. (E.M.E.U. Ans. 151-393 m) 13. The base AB is to be measured using a subtense bar of length b and the double extension layout shown in the figure. If the standard error of each of the two measured angles is + 8a develop a formula for the proportional standard error of the base length. Find the ratio a^ : a 2 which will give the minimum proportional standard error of the base length. What assumptions have you made in arriving at your answers? (R.I. C.S.) 404 SURVEYING PROBLEMS AND SOLUTIONS 14. (a) Describe, with the aid of sketches, the principles of sub- tense bar tacheometry. (b) The sketch shows two adjacent lines of a traverse AB and BD with a common sub-base BC. Calculate the lengths of the traverse lines from the following data: Angles BAC = 5°10'30" CBA = 68°56'10" YBX = 1°56'00" CDB = 12°54'20" DBC = 73°18'40" Length of bar 2 metres. (E.M.E.U. Ans. AB, 631-96 m; BD, 264-78 m) vi Y Not to scale X C Y Fig. 7.24 15. Describe the 'single bay* and 'double bay' methods of measuring linear distance by use of the subtense bar. Show that for a subtense bar L = — cot cf>/2 where L = the horizontal distance between stations, S = length of subtense bar, <f> = angle subtended by targets at the theodolite. Thereafter show that if S = 2 metres and an error of ± A<£ is made in the measurement of the subtended angle, then AL A0L L = ~ 2 where AL is the corresponding error in the computed length. Assuming an error of ± lsec(A<£) in the measurement of the sub- tended angle what will be the fractional error at the following lengths? (a) 50 metres (b) 100 metres (c) 500 metres. (N.U. Ans. (a) 1/8250; (b) 1/4125; (c) 1/825.) TACHEOMETRY 405 Exercises 7(c) (General) 16. The following readings were taken with a theodolite set over a station A, on to a staff held vertically on two points B and C. Inst. St. Horizontal Circle Vertical Circle Stadia Readings Staff Reading Reading U M L St. A 33°59'55" +10°48' 8-44 6-25 4-06 B A 209°55'21" - 4°05' 7-78 6-95 6-12 C If the instrumental constant is 100 and there is no additive con- stant; calculate the horizontal distance BC and the difference in elevation between B and C. (E.M.E.U. Ans. 587*48 ft; 93*11 ft) 17. Readings were taken on a vertical staff held at points A, B, and C with a tacheometer whose constants are 100 and 0. If the horizon- tal distances from instrument to staff were respectively 153, 212, and 298 ft, and the vertical angles +5°, +6° and -5°, calculate the staff intercepts. If the middle-hair reading was 7*00 ft in each case what was the difference in level between A, B and C? (L.U. Ans. 7-77/7-00/6-23; 8-07/7-00/5-93; 8-50/7-00/5-50; A-B. + 8-88; B-C. -48-30) 18. A theodolite has a tacheometric multiplying constant of 100 and an additive constant of zero. When set 4*50 ft above a station B, the following readings were obtained: Station at Sight Horizontal Circle Vertical Circle Stadia Readings Top Middle Bottom A C 028°21'00" 082°03'00" 20°30' 3-80 7-64 11-40 The co-ordinates of station A are E 546-2, N 0-0 and those of B are E 546-2 N -394-7. Find the co-ordinates of C and its height above datum, if the height of station B above datum is 91-01 ft. (L.U. Ans. 1083-6 E 0-1 N; +337-17 ft) 19. The following readings were obtained in a survey with a level fitted with tacheometric webs, the constant multiplier being 100 and the additive constant zero. 406 SURVEYING PROBLEMS AND SOLUTIONS Inst, at Point Staff Readings A B.M. 207*56 1-32 2-64 3-96 B 2-37 3-81 5-25 C B 5-84 7-95 10-06 D 10-11 11-71 13-31 E 8-75 9-80 10-85 F E 11-16 13-17 15-18 T.B.M. 3-78 5-34 6-90 Subsequently the level was tested and the following readings ob- tained: t. at Point Staff Readings P Q 4-61 5-36 6-11 R 3-16 3-91 4-66 S Q 4-12 4-95 5-78 R 3-09 3-17 3-25 Find the level of the T.B.M. (L.U. Ans. 212-98 ft) 20. A theodolite was set up at P, the end of a survey line on uni- formly sloping ground and the readings taken at approximately 100 ft intervals along the line as follows: At Point Elevation Angle Stadia Readings P A 4°16' 3-66 4-16 4-66 B 4° 16' 2-45 3-46 4-47 C 5°06' 1-30 2-82 4-34 D 5°06' 5-87 7-88 9-89 E 5°06' 6-15 8-65 11-15 An error of booking was apparent when reducing the observations. Find this error, the levels of the points ABCDE and the gradient PE, if the ground level below the instrument was 104-20 O.D. and the height of the instrument 4-75. Instrument constants 100 and 0. (L.U. Ans. A 112-21, B 120-48, C 128-68, D 136-66, E 144-57; Grad. 1 in 12-28) 21. The following data were taken during a survey when stadia read- ings were taken. The levelling staff was held vertically on the stations. The height above datum of station A is 475-5 ft above Ordinance Dat- um. The multiplying factor of the instrument is 99-5 and the additive constant 1*3 ft. Assume station A to be the point of origin and cal- culate the level above Ordinance Datum of each station and the horiz- ontal distance of each line. TACHEOMETRY 407 Back Instrument Fore Instrument Horizontal Vertical Staff Readings Station Station Station Height Angle Angle Upper Middle Lower - A B 4-95 - + 5°40' 9.90 8*00 6.10 A B C 5-00 164°55' + 7°00' 8.44 6.61 4.7.8 B C D 5-10 179°50' -8°20' 9.20 7.57 5-94 (R.I.C.S. Ans. A 475-50, B 509-73, C 552-33, D 502-66; AB 375-70, BC 360-04, CD 322-23) 22. The undermentioned readings were taken with a fixed-hair tach- eometer theodolite on a vertical staff. The instrument constant is 100. Calculate the horizontal distance and difference in elevation between the two staves. Instrument Horizontal Vertical Staff Station Station Circle Circle X 33°59'55" + 10°48'00" [8-44 , | 6-25} 1 4-06 J A X 209°55'21" - 4°05'00" (7-78] 6-95 U-12J B (MQB/S Ans. 587-5 ft; 93-1 ft) 23. The undernoted readings were taken at the commencement of a tacheometric survey, the multiplying factor of the tacheometer being 100 and the additive constant 1-3 ft. Calculate the co-ordinates and reduced level of station D assum- ing A to be the point of origin and the reduced level there at 657*6 ft above datum. The azimuth of the line AB is 205° 10' . (MQB/S Ans. S 893-83 ft; W 469-90 ft; 718-54 ft) 24. The following readings were taken by a theodolite used for tach- eometry from a station B on to stations A, C and D: Sight A C D Horizontal Angle 301°10' 152° 36' 205°06' Vertical Angle Stadia Readings Top Centre Bottom -5°00' 3-48 7-61 11-74 2°30' 2-15 7-92 13-70 The line BA has a bearing of N 28°46' E and the instrument has a constant multiplier of 100 and an additive constant zero. Find the slope of the line CD and its quadrantal bearing. (L.U. Ans. 1 in 7*57; N 22° 27' 10" W) 408 SURVEYING PROBLEMS AND SOLUTIONS 25. The following observations were taken with a tacheometer (mult- iplier 100, additive constant 0) from a point A, to B and C. The distance EC was measured as 157 ft. Assuming the ground to be a plane within the triangle ABC, calculate the volume of filling required to make the area level with the highest point, assuming the sides to be supported by concrete walls. Height of instrument 4'7ft, staff held vertically. At To Staff Readings Vertical Angle A B 1-48 2-73 3-98 +7°36' C 2-08 2-82 3-56 -5°24' (L.U. Ans. 297600 ft 3 ) 26. Explain the principles underlying the measurement of a horizontal distance by means of stadia readings. Using stadia readings, the horizontal distance between two stat- ions A and B is found to be 301*7 ft. The difference in height between the two stations is 3*17 ft. Cal- culate the appropriate stadia readings, stating clearly the assumptions you have made. (R.I.C.S.) 27. (i) Derive expressions for the probable errors of determination of horizontal and vertical distances tacheometrically due to known probab- le errors ds in the measurement of stadia intercept and da in the meas- urement of the vertical angle. It may be assumed that an anallactic instrument is used and the staff is held vertically. (ii) In a series of tacheometric observations a sight is taken to the top of a building where the vertical angle is about 11°30'. If the stadia intercept is 2-52 ft and ds = +0-002 5 ft and da = ± 1' what are the probable errors in determining the horizontal distances? The tach- eometric constant is 100. (iii) Assuming that using a staff graduated to 0'Olft each stadia hair can be read correctly to the nearest 0*01 ft will such a staff be good enough to give an accuracy of 1 : 500 over distances from 100 to 500 ft, and if not what accuracy can be achieved? It may be assumed that the vertical angles will be small and errors in the vertical angle can be ignored. (R.I.C.S. Ans. (ii) ±0-24 ft; ±0-03 ft) 28. A third order traverse line AB is measured by the following method: measure angles a,, a 2 , 6 shown on the diagram; measure dis- tances AC,, AC Z , by the angles at A to a subtense bar at C, and C 2 . Two measures of AB are thus obtained, their mean being accept- ed. TACHEOMETRY 409 A 2m subtense bar is centred at C, and C 2 and oriented at right angles to AC X and AC 2 . Observed horizontal angles are as follows. Subtense angles at A: to C, = 1°05'27-1" to C 2 = l°04'30-0" 6 = 100°35'33" a, = 2°51'27"; a 2 = 2°53'55" Compute the horizontal distance AB. (R.I.C.S. Ans. 104-70 m) 29. Describe in detail how you would determine the tacheometric con- stants of a theodolite in the field. Show how the most probable values could be derived by the method of least squares. Sighted horizontally a tacheometer reads r, = 6*71 and r 3 = 8*71 on a vertical staff 361*25 ft away. The focal length of the object glass is 9 in and the distance from the object glass to the trunnion axis 6 in. Calculate the stadia interval. Given D = — s + (/+ c) i (N.U. Ans. 0-05 in.) 30. Describe the essential features of a subtense bar and show how it is used in the determination of distance by a single measurement. Allowing for a 1 second error in the measurement of the angle, cal- culate from first principles the accuracy of the measurement of 200 ft if a 2 metre subtense bar is used. Show how the accuracy of such a measurement varies with distance and outline the method by which max imum accuracy will be obtained if subtense tacheometry is used in the determination of the distance between points situated on opposite banks of a river about 600 ft wide. (I.C.E. Ans. 1 in 6 800) 31. An area of ground was surveyed with a fixed stadia hair tacheo- meter (constants 100 and 0) set up in turn at each of four stations A, B, C and D. Observations were made with the staff held vertically. A BCD A formed a closed traverse and it was found that the difference in level between these four stations as calculated from the tacheomet- ric readings would not balance. 410 SURVEYING PROBLEMS AND SOLUTIONS Later it was realised that a new altitude bubble had recently been fitted to the instrument but unfortunately had not been correctly adjust- ed. In order to determine the true differences of level between the four stations, a level known to be in perfect order was used. Fieldbook observations from both surveys are as follows: Instrument Height of Staff Vertical Stadia Readings Station Instrument (ft) Station Angle (ft) A 4-55 B + 3° 4-05 5-66 7-27 B 4-60 C + 2°20' 8-71 11-02 13-33 C 4-70 D -2°30' 3-74 5-67 7-60 D 4-50 A 2-38 4-99 7-60 Backsight Intermediate sight Foresight Remarks 11-78 - - Station A 10-30 - 5-04 Change point - 3-19 - Station B 9-84 - 1-65 Change point 3-27 - 1-69 Station C 2-83 - 14-78 Change point - 11-34 - Station D 6-41 - 9-70 Change point - - 11-57 Station A Calculate the vertical angles that would have been observed from stations A and C if the altitude bubble of the tacheometer had been in correct adjustment. Describe the procedure which should be adopted in correcting the adjustment of the altitude bubble, identifying the type of instrument for which your procedure is appropriate. (I.C.E. Ans. +2°40'; -2°50') Bibliography THOMAS, W.N., Surveying (Edward Arnold). BANNISTER, A. and RAYMOND, S., Surveying (Pitman). REDMOND, F.A., Tacheometry (Technical Press). HIGGINS, A.L., Higher Surveying (Macmillan). HODGES, D.J., Articles in Colliery Engineering, Dec. 1965 - Oct. 1966. GEISLER, M., 'Precision tacheometry with vertical subtense bar applied to topographical surveying', Survey Review, Oct. 1964. WATSON and SHADBOLT, 'The subtense bar applied to mine surveying', Chartered Surveyor, Vol. 90, No. 11, May 1958. 8 DIP AND FAULT PROBLEMS Problems on gradients take a number of different forms and may be solved graphically or trigonometrically according to the accuracy re- quired. 8.1 Definitions Let ABDE represent a plane inclined to the horizontal at a°, Fig. 8.1. Level line (strike) D S JC A B Fig. 8.1 Dip. The dip of a bed, seam or road in any direction is the angle of inclination from the horizontal plane. It may be expressed as: (a) An angle from the horizontal, e.g. 6°03' , (b) A gradient, 1 vertical in n horizontal (the fraction 1/n repre- sents the tangent of the angle of inclination, whilst n represents its cotangent) or (c) A vertical fall of x inches per horizontal yard, e.g. 3 inches per yard. N.B. The term rise denotes the opposite of dip. Full Dip (or true dip) is the maximum inclination of any plane from the horizontal and its direction is always at right-angles to the mini- mum inclination (i.e. nil) or level line known as strike. In Fig. 8.1, lines AE and BD are lines of full dip, whilst ED and AB are level lines or strike lines. Apparent Dip is the dip observed in any other direction. It is always less than full dip and more than strike. In Fig. 8.1 the line AD is an 411 412 SURVEYING PROBLEMS AND SOLUTIONS apparent dip inclined at. an angle of /8° in a direction 6° from full dip. Depth of Strata. The depth of a stratum is generally measured relative to the surface, to Ordnance Mean Sea Level (as used in levelling) or, in order to obtain positive values, may be expressed relative to some arbitrary datum, e.g. the N.C.B. (National Coal Board) datum, which is 10 000 ft below M.S.L. Thickness of Strata. The true thickness is measured at right- angles to the bedding plane. For inclined strata penetrated by vertical bore- holes, an apparent thickness would be derived from the borehole core. Example 8.1 A vertical borehole passes through a seam which is known to dip at 40°. If the apparent thickness of the seam as shown by the borehole core is 5 ft calculate: (a) the true thickness of the seam; (b) the gradient of the seam. Fig. 8.2 (a) True thickness t = 5 cos 40° = 3*83 ft = 3 ft 10 in. (b) Gradient of seam cot 40? = M92 ••• Gradient is 1 in 1*192 Example 8.2 If a seam dips at 1 in 4 what is the true area of one square mile in plan Cot0 = 4 .*. 6 = 14° 02' True length of one dipping side 1760 ~ COS0 DIP AND FAULT PROBLEMS 413 1760 0-970 15 True area yd = 176 ° 2 = 3192908 sq. yd 0-97015 ±JL ~ compared with 3097 600 sq. yd in plan. 8.2 Dip Problems 8.21 Given the rate and direction of full dip, to find the apparent dip in any other direction Let Fig. 8.3 represent the plan. Graphical Solution Draw AB and AC representing strike and full dip. Let the length AC be x units long. As the line AC £ dips at I'm x, C will be 1 unit vertically below A. Draw the strike line CE through C, which is now 1 unit below strike line AB. Fig. 8.3 Any line starting from A will then be 1 unit vertically below A when it it cuts the line CE, and its gradient will be 1 in y, where y is the length measured in the same units. Trigonometrical solution Triangle ADC is right-angled at C AC A Strike B >< ^^■7/ c " e >s^> 5 (n 3' X E C • ' D AD = COS0 ie - y = ^k?) g- 1 ) COSC7 or x - y cos# (8.2) i.e. the gradient value of full dip x = the gradient value of apparent dip y x cosine of the angle between. 8.22 Given the direction of full dip and the rate and direction of an apparent dip, to find the rate of full dip (Fig. 8.4) Graphical solution Draw directions AB and AC of full dip and apparent dip respec- tively. Let AC = y units. Draw CD perpendicular to AB through C cutting the full dip 414 SURVEYING PROBLEMS AND SOLUTIONS direction at B. Length AB = x units is the gradient equivalent of full dip. Fig. 8.4 Trigonometrical solution AB = AC cos0 i.e. x - ycos(/3-a) (8.3) i.e. cot full dip = cot apparent dip cosine angle between (8.4) or tan apparent dip = tan full dip cosine angle between (8.5) Example 8.3 The full dip of a seam is 1 in 4 N 30° E. Find the gradients of roadways driven in the seam (a) due N, (b) N 75° E, (c) due E. <o c c HI "c " — *> >/ ^i. — ^\° 6 ^v£ --*~ 8 units _ ^ Due E f^\1 in 8 N^ Graphically, Fig. 8.5 (a) AC due N 1 in 4*6 (b) AD N 75° E 1 in 5*7 (c) AE due E 1 in 8 DIP AND FAULT PROBLEMS 415 Trigonometrically, (a) AC = (b) AD = cos 30° 0-8660 cos 45° 0-7071 = 4-618 (12° 13' ) = 5-657 (10° 02' ) ^ AE = OTo = «£ =8-0 (7" 08') N.B. (a) Lines at 45° to full dip have gradients 1 in yJ2x, e.g. 1 in y/2 x 4 = 1 in 5-657 (b) Lines at 60° to full dip have double the gradient value, i.e. 1 in 2x, e.g. 1 in 2x4 = 1 in 8 8.23 Given the rate and direction of full dip, to find the bearing of an apparent dip (Fig. 8.6) Fig. 8.6 This is the converse of 8.22 but it should be noted that there are two directions in which a given apparent dip occurs. Graphical solution Plot full dip, direction a°, of length x units as before. Draw strike lines through A and B. With centre A draw arcs of length y to cut strike line through B, giving 6° on either side of AB as at AC and AD. Trigonometrical solution a x cosO = — y 416 SURVEYING PROBLEMS AND SOLUTIONS Bearing AC = a-d AD = a+0 Example 8.4 A seam dips 1 in 5 in a direction 208° 00' . In what direction will a gradient of 1 in 8 occur? By plotting AC 156*45' AD 259° 15' Trigonometrical ly .". Bearing AC AD Fig. 8.7 6 = cos -1 5/8 = 51° 19' = 208° 00' - 51° 19' = 156° 41' = 208° 00' + 51° 19' = 259° 19' 8.24 Given two apparent dips, to find the rate and direction of full dip (Fig. 8.8) Fig. 8.8 DIP AND FAULT PROBLEMS 417 Graphical Solution Plot direction of apparent dips AC and AD of length y and z units respectively. Join CD. Draw AB perpendicular to CD. Measure AB in the same units as y and z. Trigonometrical solution Fig. 8.9 In triangle ADC, Fig. 8.9, AC = y AD = z DAC = 6 C ~D z-y tan 180 - $ Tan — - — = — = - 2 z + y 2 From this, angles C and are known and thus A. Triangle ABC may now be solved AB = AC cos A Bearing AB = Bearing AC - A Alternative solution also x = y cos A x = z cos(0 - A) y cos A = z cos(0 - A) Dividing by cos A = z (cos 6 cos A + sin 6 sin A) y = z (cos $ + sin 6 tan A) 418 SURVEYING PROBLEMS AND SOLUTIONS Thus tanA = — 1 22£* z sin0 sin0 = ycosecfl _ cQte z If y and z are given as angles of inclination, y = cot a z = cot )6 then tanA = tan/3 cosectf cot a- cot0 (8.6) This gives the direction of full dip. The amount x = cot a cos A or cotS = cotacosA (8.7) or tana = tan 8 cos A (8.8) N.B. If 6 = 90°, by Eq.(8.6), tanA = tan/3 cot a (8.9) Alternative solution given angles of inclination, Fig. 8.10 a. Fig. 8.10 Let A, B and C be three points in a plane with B at the highest level and A at the lowest. Let AB } C^D be a horizontal plane through A, with B, and C, vertically below B and C respectively and D the intersection on this plane of BC and B, C, each produced . AB^C^ represents the plan of points A,B and C. In the plane ABCD, AD is in the horizontal plane and is therefore a line of strike. BE is perpendicular to AD and is therefore the line of full dip. In the right-angled triangle ABB V BB, = AB, tana. In the right-angled triangle DB, B, BB, = DB, tan/3 AB y tana = DB y tan/3 DIP AND FAULT PROBLEMS 419 AB. i.e. DB, Also in triangle AB X D AB, DB. tanjS tana sine K but sin^zS sine _ tan/6 _ sin<£ tana e + 0= 180 - 6 = P e = P-cf> sin(P - cf>) = sin<£ i.e. K sin<£ cot<£ (8.10) = sin P cos (f> - cos P sin <f> = sin P cot <f> - cos P = sin(180 - 0) cot <£ - cos(180 - 0) = sin cot + cos K-cosO tanfl . — z — = t . ~ - cot0 sine/ tan a sin The value of <£ will then give the direction of the strike with full dip at 90° to this. The inclination of full dip (8) is now required. Let AB t = cot a then B, E = cot 5 as before therefore cot 8 = cot a sin 4> (8.11) i.e. Eq.(8.7) cotS = cotacosA. (A = 90 - <£) Example 8.5 A roadway dips 1 in 4 in a direction 085° 30' , inter- sects another dipping 1 in 6, 354° 30'. Find the rate and direction of full dip. Fig. 8.11 420 SURVEYING PROBLEMS AND SOLUTIONS Graphically Full dip 051° 00' 1 in 3*3 Trigonometrically C -D 6 - 4 . 180 - (360° - 354° 30' + 085° 30' ) tan = tan 7z 2 6+4 2 2^ tan 89° 00' "10 2 = I tan 44° 30' C ' D = 11° 07' 10" 2 n = 44° 30' 00" C + D Mom'nn" 2 C = 55° 37' 10" A = 34° 22' 50" Bearing of full dip = 085° 30' - (90° - 55° 37' 10") = 051° 07' 10" (AB) = 4 sin 55° 37' 10" = 4x0-825 31 = 3-30124 Gradient of full dip = 1 in 3*3 Alternative solution Gradient 1 in 4 = 14° 02' 10" Gradient 1 in 6 = 9° 27' 40" From Eq. (8.6), * x _ tan9°27'40 w _ cot91 o 00 * tanA " tanl4°02'10"sin91°00" 0-166 64 nni t^ = 0-25000 x 0-99985 + °-° 1746 = 0-666 66 + 0-017 46 = 0-68413 A = 34° 22' 40" Bearing of full dip = 085° 30' 00" - 34° 22' 40" = 051° 07' 20" Rate of full dip x = cot 8 = cot 14° 02' 10" cos 34° 22' 40" = 4 x 0-825 33 DIP AND FAULT PROBLEMS 421 = 3-301 (Gradient 1 in 3*3) 8 = 16° 51' 10" 8.25 Given the rate of full dip and the rate and direction of an appar- ent dip, to find the direction of full dip (Fig. 8.12) Fig. 8.12 There are two possible solutions. Graphical solution Draw apparent dip AC in direction and of y units. Bisect AC and draw arcs of radius l / 2 y. Through A draw arcs of length x to cut circle at B, and B 2 This gives the two possible solutions /IS, and AB Z Trigonometrical solution In triangle AB } AB, = x AO = OB, = l / 2 y 1 = /(s-x)(s-y 2 v) = f 2 V s(s-y 2 y) V where s = y 2 [ x + y 2 y + l / 2 y] = l / 2 (x + y) 2 V {x + y) Bearing of full dip = £ ± 6 (8.12) 422 SURVEYING PROBLEMS AND SOLUTIONS Example 8.6 A roadway in a seam of coal dips at 1 in 8, 125° 30', Full dip is known to be 1 in 3. Find its direction. ByEq.(8.12), tan ■/ Fig. 8.13 (8-3) 2 aJ (8 + 3) = 0-674 20 5_ 11 e = 33° 59' 20" 6 = 67° 58' 40" Bearing of full dip = 125° 30' 00" = 57° 31' 20" 67° 58' 40" or = 125° 30' 00" + 67° 58' 40" = 193° 28' 40" 8.26 Given the levels and relative positions of three points in a plane (bed or seam) , to find the direction and rate of full dip This type of problem is similar to 8.24 but the apparent dips have to be obtained from the information given. To illustrate the methods Draw an equilateral triangle ABC pf sides 600 ft each. Example 8.7 If the levels of A, B and C are 200, 300 and 275 ft respectively, find the rate and direction of full dip. DIP AND FAULT PROBLEMS 423 Fig. 8.14 Select the highest point, i.e. B — gradients will then dip away from B. Semi-graphic method As B is the highest point and A is the lowest the level of C must be between them. Difference in level AB = 300 - 200 = 100 ft Difference in level CB = 300 - 275 = 25 ft 25 .*. Level of C must be -r^r- of distance AB from B. .'. Gradient of full dip = 25 ft in 144 ft (scaled value) = 1 in 5-76 Direction scaled from plan 14° E of line AB. N.B. Strike or contour lines in the plane may be drawn parallel as shown. Alternative method Gradient B - A i.e. Gradient B-C = (300 - 200) ft in 600 ft = 1 in 6 = (300 - 275) ft in 600 ft = 1 in 24. Lay off units of 6 and 24 as in previous method to give x and y, Fig. 8. 15. Line XY is now the strike line and BZ perpendicular to XY produced is the direction of full dip. Length BZ represents the relative full dip gradient value. 424 SURVEYING PROBLEMS AND SOLUTIONS S*£i ^-14° / ^ — ■—" 6 units * A 5-8 units c - """x 24 units fl Fig. 8.15 By calculation: In triangle XYB, tan V = 24 + 6 tan60 ° _ 18x^732 . ,.0392 Y ~ X = 43° 54' Y t X - 60°00' Y = 103° 54' In triangle YZB, ZB = 6 sin 103° 54' = 6 x 0-970 72 = 5-82 (Gradient of full dip). Angle YZB = 103° 54'- 90° = 13° 54' E of line AB. Example 8.8 From the following information, as proved by boreholes, calculate the direction and rate of full dip of a seam, assuming it to be regular. Borehole Level of surface above Ordnance Datum (ft) Depth from surface to floor of seam (ft) 370 225 255 1050 405 185 Borehole B is 1414 yards from A in the direction 340° Borehole C is 1750 yards from A in the direction 264° DIP AND FAULT PROBLEMS 425 Fig. 8.16 Gradient At A, Surface + 370 Depth - 1050 Seam level - 680 ft At B, Surface + 225 Depth - 405 Seam level — 180 ft At C, Surface + 225 Depth - 185 Seam level + 70 ft CA = (680 + 70) ft in 3 x 1750 ft i.e. 1 in 7. BA = (680 - 180) ft in 3 x 1414 ft i.e. 1 in 8-484 .-. Let C,4 = 7 units and B^A = 8*484 units. In triangle AB % C, Angle A = 340 - 264 = 76° 2 tan 8-484 - 7 180 - 7 6 8-484 + 7 tan 2 1-484 tan 52° 15-484 = 0-12267 Cl I gl = 7° 00', Cl + Bl = 52° 00' 2 2 ••• C, = 59° 00' 426 SURVEYING PROBLEMS AND SOLUTIONS In triangle AC^X, XA is full dip perpendicular to C, B, ••• XA = C,A sinC, = 7 sin59° = 6 full dip is 1 in 6. Bearing AC, = 264° Bearing AX = 264° + (90 - 59) = 295 c Bearing of full dip is XA = 115° Example 8.9 Three boreholes A, B and C are put down to prove a coal seam. The depths from a level surface are 735 ft, 1050 ft, and 900 ft respectively. The line AB is N 10° E a distance of 1200 ft, whilst AC is N55°W, 900 ft. Calculate the amount and direction of full dip. B is the lowest and A is the highest. 6 = 55 + 10 = 65° 1200 1200 1050 - 735 315 = 3-809 52 a = 14° 42' 30" r _ 900 _ 900 COt P ~ 900 - 735 ~ 165 = 5-45454 jS = 10° 23' 20" 4-735 Fig. 8.17 Then by Eq.(8.10) cot<£ = tanjS tana sin0 0-18333 - cot0 = 0-26250 x 0-906 31 "°' 46631 = 0-770 62-0-466 31 =0-304 31 <f> = 73° 04' 30" Bearing of full dip = 010° - (90 - 73° 04' 30") = 353° 04' 30" = N6°55'30"W DIP AND FAULT PROBLEMS 427 Amount of full dip cot 8 = cot a sin ^ = 3-809 52 x 0-95669 = 3-644 53 i.e. Gradient 1 in 3*64 Inclination (8) = 15° 20' 40" 8.3 Problems in which the Inclinations are Expressed as Angles and a Graphical Solution is Required To illustrate the processes graphical solutions of the previous examples are given. 8.31 Given the inclination and direction of full dip, to find the rate of apparent dip in a given direction Example 8.10 Full dip N 30° E 1 in 4 (14° 02') Apparent dip (a) Due N, (b) N 75° E. Apparent dips (a) 12*15' (b) 10*00' 10*00' Fig. 8.18 Draw AB (full dip) N 30° E of convenient length say 3 in. Through A and B draw strike lines AX and BY and assume that AX is 1 unit vertically above BY. At B set off the inclination of full dip (i.e. 14° 02') to cut AX at B,. AB^B may now be considered as a vertical section with AB A of length 1 unit. Draw a circle of centre A, radius AB^ (1 unit). Now draw the direction of apparent dips AC (due N) to cut strike BY at C and AD (N 75° E) to cut strike BY at D. 428 SURVEYING PROBLEMS AND SOLUTIONS Also draw AC, and AD, perpendicular to AC and AD respectfully, cutting the circle at C, and D,. N.B. AB, = AC, = AD, =1 unit. AC Then -^ represents the gradient of AC (1 in 4*6). The angle ACC, is the angle of dip 12° 15'. AD Similarly -^j- represents the gradient of AD (1 in 5-7). The angle ADD, is the angle of dip 10°00' . 8.32 Given the inclination and direction of full dip, to find the direc- tion of a given apparent dip Example 8.11 Full dip 1 in 5 (11°18') 208°00'. Apparent dip 1 in 8 (7°07'). 90*- Direction of apparent dip 259° 30' and 156*30' Fig. 8.19 Draw AB in direction as before of say 3 in. At A and B draw the strike lines AX and BY. At B set off the angle of full dip 11°18' to cut AX produced at A, . Draw a circle of centre A and radius AA^ Produce BA to cut the circle at Q and set off the angle (90° - 7°07 ; ) to cut AX at P— this represents a section of the apparent dip, AP being the length of the section proportional to the vertical fall of 1 unit G4Q). With centre A and radius AP draw an arc to cut BY at C, and C 2 , i.e. AC, = AC Z = AP. These represent the direction of the apparent dips required. DIP AND FAULT PROBLEMS 429 8.33 Given the inclination and direction of two apparent dips to find the inclination and direction of full dip Example 8.12 1 in 4 (14°) 085° 30' 1 in 6 (9°30') 354° 30' Direction of full dip 051* 00"! Amount " " " 17' Fig. 8.20 Draw AC and AD in the direction of the apparent dips. With A as centre draw a circle of unit radius. Draw AC^ and AD^ perpendicular to AC and AD respectively. Set off at C, (90° - 14°) and at D (90° - 9°30'). This will give 14° at C and 9°30' at D. Join CD, the strike line. Draw AB perpendicular to CD through A and AB^ perpen- dicular to AB. Join B, B and measure the direction of AB and the angle of inclination B^BA. Graphical solution: Full dip 17 c 051°00'. Exercises 8(a) 1. The full dip of a seam is 4 inches in the yard. Calculate the angle included between full dip and an apparent dip of 3 inches in the yard. (Ans. 41° 24') 2. The angle included between the directions of full dip and apparent dip is 60°. If the apparent dip is 9°10', calculate the full dip, expres- sing the answer in angular measure, and also as a gradient. (Ans. 17° 50'; 1 in 3*1) 3. On a hill sloping at 18° runs a track at an angle of 50° with the line of greatest slope. Calculate the inclination of the track and also its length, if the height of the hill is 1500 ft. (Ans. 11° 48'; 7335 ft) 430 SURVEYING PROBLEMS AND SOLUTIONS 4. The full dip of a seam is 1 in 3 in a direction N 85° 14' W. A road- way is to be driven in the seam in a southerly direction dipping at 1 in 10. Calculate the quadrant and azimuth bearings of the roadway. (Ans. S 22°14'W; 202° 14') 5. The full dip of a seam is 1 in 5, N 4° W. A roadway is to be set off rising at 1 in 8. Calculate the alternative quadrant bearings of the roadway. (Ans. S 47° 19' W; S 55°19' E.) 6. The following are the particulars of 3 boreholes. Borehole Surface Level A.O.D. Depth of Borehole A 600 ft 500 ft B 400 ft 600 ft C 1000 ft 600 ft If the distance from A to B is 1200 ft and from B to C 1800 ft, calculate the gradiants of the lines AB and BC. (Ans. 1 in 4; 1 in 3) 7. A and B are two boreholes which have been put down to prove a seam. They are on the line of full dip of the seam, the direction of line BA being N 50°E and its plan length 1000 ft. Borehole Surface Level Depth A 600 ft 750 ft B 800 ft 700 ft A shaft is to be sunk at a point C, the surface level of C being 1000 ft, the length BC 800 ft, and the direction of BC due East. Calculate (a) the dip of the seam from B to C, (b) the depth of the shaft at C. (Ans. (a) 1 in 5- 23; (b) 1053 ft) 8. A seam dips at 1 in 12-75, S 17° W and at 1 in 12*41, S 20°15' E Calculate the magnitude and direction of full dip. (Ans. 1 in 11-64; S 6°46' E). 9. The co-ordinates and level values of points A, B and C respec- tively, in a mine, are as follows: Departure (ft) Latitude (ft) A +119-0 + 74-0 9872 B -250-0 + 787-5 9703 C +812-0 +1011-0 9805 (a) Plot the positions of the points to a convenient scale and graphic- ally determine the direction and amount of full dip. Levels in ft above a datum 10000 ft below O.D. DIP AND FAULT PROBLEMS 431 (b) Calculate the direction and amount of full dip. (Ans. (a) N 38° W; 1 in 4-66. (b) N 37°56' W; 1 in 4-672) 10. In a steep seam a roadway AB has an azimuth of 190° dipping at 22° and a roadway AC has an azimuth of 351° rising at 16°. Calculate the direction and rate of full dip of the seam. (Ans. 225° 54' 50", 26° 31') 11. A cross-measures drift, driven due South and dipping at 16°, passes through a bed of shale dipping due North at 27°. The distance between the roof and floor of the bed of shale measured on the floor of the drift is 56 ft. Calculate the true and vertical thickness of the bed. (Ans. 38-19 ft; 42-86 ft) 12. A small colliery leasehold was proved by 3 bores. The surface level at each bore and the depth to the seam were as follows: Bore Surface Level Depth A 30 ft above O.D. 190 ft B 20 ft above O.D. 220 ft C 10 ft above O.D. 240 ft Bore B is 560 yds from A N 30° E and bore C is 420 yd from B S 60° E. Find graphically, or otherwise, the direction of strike and the rate of full dip in inches per yard. If a shaft is sunk 800 yd from A in a direction S 45° E at what depth below datum will it reach the seam ? (Ans. N 15° W; 1% in. per yd; 200-4 ft)- 13. An underground roadway driven on the strike of the seam has a bearing S 30° E. The seam has a full dip of 8 in to the yd in a northerly direction. At a point A on the roadway another road is to be set off rising at 1 in 5*8. Calculate the alternative bearings on which this second road may be set off. (Ans. N 80° 54' W; S 20° 54' E) 14. Three bores A, B and C have been put down to a coal seam. B is due north from A, 1000 ft, and C is N 76° W, 850 ft from A. The surface levels of the boreholes are the same. The depth of A is 700ft, of B 1250 ft and of C 950 ft. Calculate the direction and rate of full dip and the slope area in the triangle formed by the boreholes. (Ans. N 16° 49' W; 1 in 1-74; 475610 sq ft) 432 SURVEYING PROBLEMS AND SOLUTIONS 8.4 The Rate of Approach Method for Convergent Lines In Fig. 8.21, let AC rise from A at 1 in K. AD dip from A at 1 in M. AB be the horizontal line through A. Fig. 8-21 AF = 1 unit then GF = y units and FE = -T-. units M Comparing similar triangles ADC and AEG, CD GE ± 1_ BA FA KM ■'■ CD = BA {\ + h) and BA = CD I 1 K + M Thus if 2 convergent lines CA and DA are CD vertically apart, the horizontal distance BA when they meet _ CD ~ I I ( 8 ' 13 > W/zen tfie fines both dip from A (Fig. 8.22) B F A it i c M "ril S^ M-K\ \ \/ ^^e Fig. 8 DIP AND FAULT PROBLEMS 433 Let AC dip from A at 1 in K AD dip from A at 1 in M. As before, CD BA G£ FA CD and BA = _1_ _ 1 M K 1 BA CD £_ 1 \M K) Therefore, if the two convergent lines CA and DA are CD verti- cally apart, the horizontal distance BA when they meet CD 1 1 M~ K (8.14) Thus if a rise is considered +ve and dip -ve, then the general expression applies. Horizontal distance = Yggical distance apart 1_^ j_ M " K (8.15) Example 8.13 Two seams of coal, 100 ft vertically apart, dip at 1 in 6. Find the length of a (drift) roadway driven between the seams (a) at a rise of 1 in 4 from the lower to the upper and (b) at a fall of 1 in 2 from upper to lower. JlPPer searn . > ft L i ■ s eorrj T 100' ^».--"~~~~ " " — — — — Li2_e "^-^ c, c (a) In triangle ADB, Fig. 8.23 AD falls at 1 in 6. BD rises at 1 in 4. AB = 100 ft 434 SURVEYING PROBLEMS AND SOLUTIONS Then, by Eq. (8.15), D,D = T AB -(4) i.e. plan length of drift inclined length of drift 100 1 1 4 + 6 12 x 100 3+2 = 240 ft 240 x V4 2 + l 2 4 240 x yl7 4 = 247*2 ft (b) Similarly, by Eq.(8.15), AB C. C plan length 100 6 \ 2) 2 6 C^C = ^ = 300 ft inclined length AC = 3 °° X ^ 5 = 336'0 ft Example 8.14 Two parallel levels, 200 ft apart, run due East and West in a seam which dips due North at 3 in. to the yard. At a point A in the lower level a cross-measures drift rising at 6 in. to the yard and bearing N 30° E is driven to intersect another seam, situated 200 ft vertically above the seam first mentioned, at a point C. From a point DIP AND FAULT PROBLEMS 435 B in the upper level due South from A another cross-measures drift rising at 12 in. to the yard and bearing N 30° E is also driven to inter- sect the upper seam at a point D. Calculate the length and azimuth of CD. (M.Q.B.) (418-70, 725-20) 21 227-13) (0,-200) Fig. 8.25 Apparent dip of seam in direction of drift (Fig, 8.25) ByEq(84 > .,._,„ cot full dip cosine angled between = X * = 13-856. cos 30° To find length of drift AC, Fig. 8.26(a), Plan length AC 200 200 x 83-136 1 1 19-856 6 ' 13-856 = 837-39 ft 436 SURVEYING PROBLEMS AND SOLUTIONS To find length of drift BD, Fig. 8.26(b), Plan length BD = 200 200 x 41-568 1 1 16-856 3 + 13-856 = 493-21 ft (a) 200' UP 13-856 837-39 A '^-?--'" Fig. 8.26 Assuming the two levels are 200 ft apart in plan, the relative positions of A, B, C and D can now be found from the co-ordinates (see chapter 3). AC N 30° E 837-39 ft .-. P.D. = +418-70 .'. P.L. = +725-20 Total departure of C = + 418-70 Total latitude of C = +725-20 sin 30° = 0-50000 cos 30° = 0-86603 (relative to A) AB due South 200 ft BD N 30° E 493-21 ft sin 30° = 0-50000 cos 30°= 0-86603 Total departure of B = 0*0 Total latitude of B = -200-0 .-. P.D. = +246-21 .-. P.L. = +427-13 Total departure of D = +246-21 DIP AND FAULT PROBLEMS 437 Total latitude of D = +427-13 - 200 = +227-13 « • n- rn ♦ -i 246-21 - 418-70 Beanng of line CD = tan 22y . 13 _ ^.^ -i -172-49 " tan -498-07 = tan -1 0-346 32 = S 19° 06' 10" W 498-07 498-07 Length of line CD = cosl9 o 06 / l0 « = q-94493 = 527-10 ft (horizontal length) If the inclined length is required, AC rises at 1 in 6 •'• the difference in level A - C = + ™^- = +139-56 ft 6 BD rises at 1 in 3 •*• the difference in level 493-21 B - D = +^\^ = +164-40 ft .-. level of D relative to B = +164*40. AB rises at 1 in 12 •*. the difference in level A-B = + ^ = +16-67 ft /. Level of D relative to C = +164-40 + 16*67 - 139-56 = +41-51 ft. .-. Inclined length CD = ./(527-10 2 + 41-51 2 ) = V( 277 834 + 1723) = 528-7 ft 8.5 Fault Problems 8.51 Definitions A geological fault is a fracture in the strata due to strains and stresses within the earth's crust, accompanied by dislocation of strata. The direction of movement decides the nature of the fault. With simple displacement either the corresponding strata are forced apart giving a normal fault or movement in the opposite direction causes an overlap known as a reverse fault. Many variations are possible with these basic forms. 438 SURVEYING PROBLEMS AND SOLUTIONS MAM////////. Normal fault Reverse fault Fig. 8.27 Fig. 8.27 only illustrates the end view and no indication is given of movement in any other direction. The following terms are used (see Fig. 8.27): FF % is known as the fault plane, B is the upthrow side of the fault, C is the downthrow side of the fault, 6 is the angle of hade of the fault (measured from the ver- tical), BE is the vertical displacement or throw of the fault, EC is the horizontal displacement, lateral shift or heave, causing an area of want or barren ground in the normal fault. Faults which strike parallel to the strike of the bed are known as strike faults. Faults which strike parallel to the dip of the beds are known as dip faults. Faults which strike parallel to neither dip nor strike are known as oblique faults and are probably the most common form, frequently with rotation along the fault plane, Fig. 8.28 . ab — strike slip be = dip slip ac = net or resultant slip bd = vertical throw Fig. 8.28 Diagonal or oblique fault DIP AND FAULT PROBLEMS 439 Where the direction and amount of full dip remain the same on both sides of the fault, the fault is of the simple type and the lines of con- tact between seam and fault on both sides of the fault are parallel. Where the direction and/or the amount of dip changes, rotation of the strata has taken place and the lines of contact will converge and diverge. The vertical throw diminishes towards the convergence until there is a change in the direction of the throw which then increases, Fig. 8.29. Down throw Down throw Fig. 8.29 N.B. The strike or level line of a fault is its true bearing, which will differ from the bearing of the line of contact between seam and fault plane. Example 8- 15 A vertical shaft, which is being sunk with an excavated diameter of 23 ft 6 in. passes through a well-defined fault of uniform direction and hade. Depths to the fault plane below a convenient horizontal plane are taken vertically at the extremities of two diameters AB and CD, which bear north-south and east-west, respectively. The undernoted depths were measured: at A (north point) 10' 1" at B (south point) 26' 3" at C (east point) 4' 0" at D (west point) 32' 4" Calculate the direction of the throw of the fault and the amount of hade. Express the latter to the nearest degree of inclination from the vertical. (M.Q.B./S) N (10-1") (-32'-4")W Fig. 8.30 E (4'-0") S(-26 -3 ) 440 SURVEYING PROBLEMS AND SOLUTIONS Fig. 8.31 Gradient NS line (26'3"-10'l") in 23' 6" i.e. 16'2"in23'6" = 1 in 1-453 57 .-. j3 = 34°31'40" Gradient EW line (32'4"-4'-0") in 23'-6" i.e. 28'-4"in 23'6" * =1 in 0-82951 a = 50° 19' 30" From Eq. (8.6), tan A = tan /3 cosec 6 cot a - cot 6 but 6 = 90° a tan A = tan£ cot a (by Eq. 8.9) = tan 34°31'40" x 0-829 51 = 0-570 70 8 = 29°42'50" •*. Bearing of full dip = 270° - 29°42'50" = 24 0° 17' 10" From Eq.(8.7), cot 8 = cot a cos A = 0-829 51 cos29°42'50" = 0-720 44 8 = 54° 13' 50" /. Angle of hade = 90 - 8 = 35°46' 10" (36° to nearest degree) Example 8- 16 A roadway advancing due West in a level seam encoun- ters a normal fault running North and South, with a hade of 30°, which DIP AND FAULT PROBLEMS 441 throws the seam up by a vertical displacement of 25 ft. A drift rising at 1 in 3*6 is driven from the point where the road- way meets the fault on the East side to intersect the seam on the West side of the fault. Find, by drawing to scale, the inclined length of the drift and check your answer by calculation. Fig. 8.32 Length of drift scaled from drawing = 93*5 ft. BC 25 Length of drift but AB = sin (7 cot(9 = 3*6 = 15°31' 25 AB Length of drift sinl5°31 = 93-4 ft. id = 93-42 ft. Example 8.17 A roadway dipping 1 in 8 in the direction of full dip of a seam strikes an upthrow fault, bearing at right-angles thereto. Following the fault plane a distance of 45 ft the seam is again located and the hade of the fault proved to be 30°. Calculate the length of a cross-measures drift to win the seam, commencing at the lower side of the fault and rising at 1 in 6 in the same direction as the roadway. Fig. 8.33 Vertical throw of fault FB = 45 cos 30° = 38*97 ft Lateral displacement FC = 45 sin 30 = 22' 50 ft 442 SURVEYING PROBLEMS AND SOLUTIONS In triangle EFC EF = EC 22-50 = 2-81 ft 8 8 .-. EB = FB + EF = 38-97 + 2-81 = 41-78 ft. To find the plan length of the drift by the rate of approach method, BG = 41-78 1 1 8 + 6 = 143-24 ft Inclined length of drift BD = 143-24 x 41-78 x 48 14 = 145-22 ft Example 8.18 A roadway, advancing due East in the direction of full dip of 1 in 8, meets a downthrow fault bearing S 35° E, with a throw of 60 ft and a hade of 30°. At a distance 150 ft along the roadway, back from the fault, a drift is to be driven in the same direction as the roadway in such a way that it meets the point of contact of seam and fault on the downthrow side. Calculate the inclined length and gradient of the drift, assuming the seam is constant on both sides of the fault. Section on fault 60' Fig. 8.34 DIP AND FAULT PROBLEMS 443 In plan Width of barren ground = BX = 60 tan 30° = 34-64 ft. On the line of the roadway BC = 34-64 sec 35° = 42- 29 ft (= QC). As X is 60 ft vertically below B, it is necessary to obtain the relative level of C. XC = BC sin 35° XY = XC sin 35 = BC sin 2 35° = 42- 29 sin 2 35° = 13-91 ft As the dip of XY is 1 in 8, the level of Y relative to X is O But YC is the line of strike .'. Level of Y = level of C = 1-74 ft below X = 60 + 1-74 ft below A = 61-74 ft below A (i.e. BQ = MP). In section Gradient of roadway AB = 1 in 8. .-. cot0 = 8 = 7°08' Thus AM = 150sin7°08' = 18-63 ft and MB = 150 cos7°08' = 148-84 ft (= PC). Difference in level A - C = AP = AM + MP = 18-63 + 61-74 = 80-37 ft Plan length of drift AC = PC = PQ + QC = 148-84 + 42-29 = 191-13 ft ••• Gradient of drift = 80-37 ft in 191-13 ft , . 191-13 = luk ~WW = 1 in 2-378 cot<£ = 2-378 <f> = 22°49' Length of drift = 19 1- 13 sec 22°49 ' = 207-36 ft 8.52 To find the relationship between the true and apparent bearings of a fault The true bearing of a fault is the bearing of its strike or level line. The apparent bearing of a fault is the bearing of the line of con- 444 SURVEYING PROBLEMS AND SOLUTIONS tact between the seam and the fault. The following conditions may exist: (1) If the seam is level, the contact line is the true bearing of the fault. (2) The apparent bearings are alike, i.e. parallel, if the full dip of the seam is constant in direction and amount. N.B. The throw of the fault will also be constant throughout its length— this is unusual. (3) The apparent bearings differ, due to variation in direction or amount of full dip on either side of the fault. The barren ground will thus diminish in one direction. N.B. The throw of the fault will vary and ultimately reduce to zero and then change from upthrow to down- throw. This is generally the result of rotation of the beds. Two general cases will therefore be considered: (1) When the throw of the fault opposes the dip of the seam. (2) When the throw of the fault is in the same general direction as the dip of the seam. Let the full dip on the downthrow side be 1 in x, the full dip on the upthrow side be 1 in y, the angle between the full dip and the line of contact be a, the angle between the line of contact and the true bearing of the fault be /3, the angle of hade be 6° from the vertical, the throw of the fault be t ft down in the direction NW, the angle between the full dip 1 in y and the true bearing of the fault be <f>. 8.53 To find the true bearing of a fault when the throw of the fault opposes the dip of the seam (Fig. 8.35) If the throw of the fault is t ft, then D will be + 1 ft above A, and for the true bearing of the fault DC, C must be at the same level as D. Angle ADC = 90° with DC tangential to the arc of radius t . tan Q. The full dip 1 in x requires a horizontal length AB = tx ft. The same applies on the other side of the fault. E must be at the same level as A, and EA, the true bearing of the fault, must be paral- lel to DC; also for FE to be the strike in the seam DF must equal ty ft. Referring to Fig. 8.35, In triangle ABC, AC = tX cos a DIP AND FAULT PROBLEMS In triangle ACD, t tan0 445 sinjS, = — = tan0 cos a, (8.16) tx cos a, In triangles ADE and EDF, but . tan0 coscl, smfl, = - y a 2 = <£ + j8 2 . tan0 •• sin P 2 - — — cos(<£+/8 2 ) (8.17) (8.18) tan0 (cos tfi cos /3 2 - sin sin /3 2 ) i.e. y cot# = cos <f> cot /3 2 - sin <f> y cot + sin <f> cotj8 2 COS0 (8.19) Hence the true bearing of the fault = bearing of contact line AC - /3, Bearing of contact line ED (8.20) = true bearing of fault + /3 2 = bearing of AC - ft, + ft 2 (8.21) Fig. 8.35 446 SURVEYING PROBLEMS AND SOLUTIONS 8.54 Given the angle 8 between the full dip of the seam and the true bearing of the fault, to find the bearing of the line of contact (Fig. 8.35) From Eq.(8.16), sin a, = 8+P, tanO cos a, x tan0 /o. o \ = cos (5 + ft,) x cot# = cos g cos ft, - sin 8 sinft, cot ft, sinp^ = cos 5 cot ft, - sin 8 x cot d + sin 8 ~ cos 8 (8.22) Example 8.19 A plan of workings in a seam dipping at a gradient of S W 1 in 3 in a direction S 30° E shows a fault bearing 45 F in the seam which throws the measures down to the North-West. The hade of the fault is 30° to the vertical. Calculate the true bearing of the fault. From Eq.(8.16), sin ft = Fig. 8.36 cos a tanfl _ cos 75° tan 30° x 3 0-149430 = 0-049 810 ft = 2°51'20" /. True bearing = 045°00'00" - 2°51'20" o r\Q> An" = 042° 08' 40 DIP AND FAULT PROBLEMS 447 Example 8.20 A seam dipping 1 in 5, S 60° E, is intersected by a fault the hade of which is 30° to the vertical and the bearing N 35° W. The fault is a downthrow to the South- West. Calculate the bearing of the fault as exposed by the seam. Fig. 8.37 As full dip and hade oppose each other, by Eq. (8.22) x cot 6 + sin 8 cot/3 = = cos 5 5 cot30 o + sin25° cos 25° 10-02181 5°42' Bearing of fault exposed in seam = 325°- 5° 42' = 319° 18' = N40°42 / W Example 8.21 Headings in a seam at A and B have made contact with a previously unlocated fault which throws the measures up 100 ft to the south-east with a true hade of 40° from the vertical. Full dip is known to be constant in direction, namely 202° 30' , but the amount of dip changes from 1 in 5 on the north side to 1 in 3 on the south side of the fault. Given the co-ordinates of A and B as follows: A 3672-46 ft E. 5873-59 ft N. B 4965-24 ft E. 7274*38 ft N. Calculate (a) the true bearing of the fault, (b) the bearings of the lines of contact between fault and seam. 448 SURVEYING PROBLEMS AND SOLUTIONS /a % / 4oV Section on fault Fig. 8.38 Co-ordinates E N A 3672-46 5873*59 B 4965-24 7274-38 dE + 1292-78 dN + 1400-79 tan bearing AB = f = ±f|§f = 0-92289 bearing of contact line AB = 042° 42' Then a, = 180° + 042°42' - 202°30' = 20°12' By Eq.(8.16), . Q cos20°12' tan 40° sm/3, = = = 0-157 50 j8, = 9°04' Bearing of line of strike of fault, i.e. true bearing of fault = 042°42' - 9°04' = 033°38' Now <fi = a,-/3, = 20 o 12'-9 o 04 , = 11°08' By Eq.(8.19) cotjS 2 = 3 cot40° + sinll o 08' cosll°08' = 3-840 62 j8 2 = 14°34' Bearing of line of contact on upthrow side of fault = 033°38' + 14°34' = 048°12' DIP AND FAULT PROBLEMS 449 8.55 To find the true bearing of a fault when the downthrow of the fault is in the same general direction as the dip of the seam (Fig. 8.39) Fig. 8.39 . a tan 6 cos a, smjS, = - before and sin j6 2 = tan# cosa.j (8.23) (8.24) but a * = <f>~Pz sin)8 2 = tan0 (cos cf> cos 2 + sin cf> sin 2 ) y cotd = cos<£ cot/S 2 + sin<£ . a y cot - sin cb cotj8 2 = ^ (8.25) (8.26) COS0 Hence true bearing of fault = bearing of contact line CA + j8, bearing of contact line DE = true bearing of fault - j8 2 = bearing of CA + ff, - /3 2 (8.27) 8.56 Given the angle 5 between the full dip of the seam and the true bearing of the fault, to find the bearing of the line of contact Here i.e. 8 = a 1+ j8, a, = 5,-0, 450 SURVEYING PROBLEMS AND SOLUTIONS then, from Eq.(8.23), sin/3, = -j- cos(5-£,) L „ x cot0 - sinS, giving cot/3, = ^ (8.28) Example 8-22 A fault is known to be an upthrow in a NW direction with a full hade of 30° to the vertical. The bearing of the fault as exposed in the seam is N 40° E, and the full dip of the seam is 1 in 5 S 35° W. Find the true bearing of the fault. Fig. 8.40 As the downthrow of the fault is in general direction as the full dip, by Eq. (8.23), . „ tan0 cos a sin0 = - tan 30° cos 5° 5 = 0-11503 B = 6°36' .-. Bearing of fault = N 40° + 6°36' E = N 46 36 y E 8.6 To Find the Bearing and Inclination of the line of Intersection (AB) of Two Inclined Planes Let (1) the horizontal angle between the lines of full dip of the two planes inclined at a and j8 respectively be S, (2) the horizontal angle between the line of full dip a and the intersection line AB of inclination <f> be 0. DIP AND FAULT PROBLEMS 451 Fig. 8.41 From Eq. (8.5), tan <f> = tan a cos 6 also = tanjS cos(S-0) tan a cos = tan B (cos 8 cos 6 + sin 5 sin 0) i.e. tan a cot j8 = cos 8 + sin 5 tan 6 „ tan a cot B - cos 8 tan0 = t 1 -^ — sin 3 (8.29) If plane (a) is a plane in the form of a seam and plane (b) is a fault plane of hade (90 - B) it a tan a tan B - cos 8 then tan0 = £-=: sino (8.30) Example 8.23 A is a point on the line of intersection of two inclined planes. The full dip of one plane is 1 in 8 in a direction 222° 15' and the full dip of the other is 1 in 4 in a direction 145°25' . If B is a point to the south of A on the line of intersection of the two planes, calculate the bearing and the inclination of the line AB. Let the angle of the full dip of plane (1) be a = cot _1 4 = 14°02' of plane (2) be B. Then the horizontal angle 8 between them = 222° 15' - 145° 25' = 76°50' From Eq.(8.29), a tana cot B - cos 8 tan0 = f— =- sin 5 i.e. tan0 = 8 tanl4°02'-cos76°50' sin76°50' 1-99960 -0-22778 0-97371 = 1-81966 452 SURVEYING PROBLEMS AND SOLUTIONS 6 = 61°12'30" .'. Bearing of AB = 145°25'+ 61°12'30" = 206°37 , 30" Angle of inclination AB(<f)) tan <f> = tan a cos 6 = 0-120 38 cf> = 6°52' i.e. 1 in 8*3 Exercises 8(b) (Faults) 15. A heading in a certain seam advancing due East and rising at 6 in. to the yd met a fault with a displacement up to the East and the seam has been recovered by a cross-measures drift rising at 18 in. to the yd in the same direction as the heading. The floor of the seam beyond the fault where the gradient is unchanged was met by the roof of the drift when it had advanced 345 ft, and the roof of the seam when it had advanced 365 ft Calculate the thickness of the coal seam. (M.Q.B./M Ans. 5 ft 10^ in.) 16. Two parallel seams 60 ft vertically apart dip due W at 1 in 6. A drift with a falling gradient of 1 in 12 is driven from the upper to the lower seam in a direction due E. Calculate the length of the drift. (Ans. Hor. 240 ft; Incl. 240-82 ft). 17. A roadway in a level seam advancing due N meets a normal fault with a hade of 30° from the vertical and bearing at right-angles to the roadway. An exploring drift is set off due N and rising 1 in 1. At a dis- tance of 41 ft, as measured on the slope of the drift, the seam on the north side of the fault is again intersected. (a) Calculate the throw of the fault and the width of the barren ground. (b) If the drift had been driven at 1 in 4 (in place of 1 in 1) what would be the throw of the fault and the width of barren ground ? (Ans. (a) 29 ft; 16-7 ft (b) 9-9 ft; 5-7 ft) 18. A roadway AB, driven on the full rise of a seam at a gradient of 1 in 10, is intersected at B by an upthrow fault, the bearing of which is parallel with the direction of the level course of the seam, with a hade of 30° from the vertical. From B a cross-measures drift has been driven in the line of AB produced, to intersect the seam at a point 190 ft above the level of B DIP AND FAULT PROBLEMS 453 and 386 ft from the upper side of the fault as measured in the seam. Calculate the amount of vertical displacement of the fault and the length and gradient of the cross-measures drift. Assume that the direc- tion and rate of dip of the seam is the same on each side of the fault. (Ans. 151-6 ft; 507-78 ft; 1 in 2-5) 19. A roadway advancing due East in a level seam meets a fault bear- ing North and South, which hades at 30°. A drift, driven up the fault plane in the same direction as the roadway, meets the seam again at a distance of 120 ft. Calculate the length and gradient of a drift rising from a point on the road 400 ft to the West of the fault which intersects the seam 100 ft East of the fault. (Ans. 570 ft; 1 in 5-4) 20. The direction and rate of full dip of two seams 60 yd vertically apart from floor to floor are N 12° E and 4 l / 2 in. to the yd respectively. Calculate the length of a cross-measures drift driven from the lower seam to intersect the upper, and bearing N 18° W. (a) if the drift is level; (b) if it rises at a gradient of 4 in. to the yd. (M.Q.B./M Ans. (a) 1662-7 ft; (b) 825-2 ft) 21. A level roadway AB bearing due West, in a seam 8 ft thick, strikes a normal fault at B, the point B being where the fault plane cuts off the seam at floor level. The hade of the fault, as measured in the roadway is 35° from the vertical. A proving drift is driven on the same bearing as the roadway and dipping from the point B at 1 in 3. The floor of the drift intersects the floor of the seam, on the lower side of the fault at a point C and BC is 80 ft measured on the slope. At C the seam is found to be rising at 1 in 10 due East towards the fault. Draw a section to a scale of 1 in. = 20 ft on the line ABC show- ing the seam on both sides of the fault, the drift BC and the fault plane and mark on your drawing the throw of the fault and the distance in the seam from C to the fault plane. (M.Q.B./M Ans. 19 ft; 63 ft) 22. The direction of full dip is due North with a gradient of 1 in 6 on the upthrow side and 1 in 9 on the downthrow side. Workings show the line of contact of the seam and fault on the upthrow side as N 45° E with a fault hade of 30° and throw of 30 ft. Calculate (a) the true bearing of the fault, (b) the bearing of the line of contact on the downthrow side. (Ans. 041°06'; 043°45') 454 SURVEYING PROBLEMS AND SOLUTIONS Exercises 8(c) (General) 23. In a seam a roadway AB on a bearing 024°00' dipping at 1 in 9 meets a second roadway AC bearing 323°00' dipping at 1 in 5 x /2. Calculate the rate and direction of full dip of the seam. (Ans. 331°13'; 1 in 5-44) 24. Two seams 60 ft vertically apart, dip at 1 in 8 due N. It is re- quired to drive a cross-measures drift in a direction N 30° W and rising at 1 in 5 from the lower to the upper seam. Calculate the length of the drivage. (Ans. 198-5 ft) 25. A roadway driven to the full dip of 1 in 12 in a coal seam meets a 240 ft downthrow fault. A cross-measures drift is set off in the di- rection of the roadway at a gradient of 6 in. to the yd. In what distance will it strike the seam again on the downthrow side if (a) the hade is 8° from the vertical, (b) the hade is vertical ? (Ans. (a) 2885-7 ft; (b) 2919-8 ft) 26. Three boreholes A, B and C intersect a seam at depths of 540 ft, 624 ft and 990 ft respectively. A is 1800 ft North of C and 2400 East of B. Calculate the rate and direction of full dip. (Ans. 1 in 3-96; S 7°58' W) 27. The rate of full dip of a seam is 4^ in. to the yd and the direc- tion is S 45° E. Find, by calculation or by drawing to a suitable scale, the inclina- tions of two roadways driven in the seam of which the azimuths are 195° and 345° respectively. (Ans. 1 in 16; 1 in 9-24) 28. Two parallel roadways AB and CD advancing due North on the strike of a seam are connected by a road BD in the seam, on a bearing N 60° E. The plan length of BD is 150 yd and the rate of dip of the seam is 1 in 5 in the direction BD. Another roadway is to be driven on the bearing N 45° E to connect the two roadways commencing at a point 200 yd out by B on the road AB, the first 20 yd to be level. Calculate the total length of the new roadway and the gradient of the inclined portion. (Ans. 186-44 yd; 1 in 5*46) 29. A roadway AB, 700 ft in length, has been driven in a seam of coal on an azimuth of 173°54'. It is required to drive a cross-measures drift from a point C in another seam at a uniform gradient to intersect at a point D, the road AB produced in direction and gradient. The DIP AND FAULT PROBLEMS 455 levels of A, B and C in feet below Ordnance Datum and the co-ordin- ates of A and C in feet are respectively as follows: Levels Latitude Departure A -1378 +9209 +18041 B -1360 C -1307 +6180 +17513 Calculate the length AD, and the length and gradient of the propo- sed drift CD, assuming that the latter is to have an azimuth of 032°27'. (Ans. 1892-9 ft; 1359-0 ft; 1 in 3-27) 30. A heading AB driven direct to the rise in a certain seam at a gradient of 6 in. to the yd and in the direction due N is intersected at B by an upthrow fault, bearing at right-angles thereto with a hade of 30° from the vertical. From the point B a cross-measures drift has been driven in the direction of AB produced, and intersects the seam at a point 420 ft from the upper side of the fault. The levels at the beginning and end of the cross-measures drift are 100 ft and 365 ft respectively above datum. Calculate (a) the vertical displacement of the fault, (b) the length and gradient of the drift. Assume the direction and rate of dip of the seam to be uniform on each side of the fault. (M.Q.B./M Ans. (a) 195-9 ft (b) 590-2; 1 in 2) 31 . A roadway in a seam dipping 1 in 7 on the line of the roadway meets a downthrow fault of 30 ft with a hade of 2 vertical to 1 horizon- tal. Calculate the length of the drift, dipping at 1 in 4 in the line of the roadway, to win the seam, a plan distance of 50 ft from the dip side of the fault; also the plan distance from the rise side of the fault from which the drift must be set off. Assume the gradient of the seam to be uniform and the line of the fault at right- angles to the roadway. (Ans. 267-97 ft; 194-97 ft) 32. A roadway, bearing due East in a seam which dips due South at 1 in 11, has struck a fault at a point A. The fault which, on this side, runs in the seam at N 10° W is found to hade at 20° from the vertical and to throw the seam down 30 ft at the point A. The dip of the seam on the lower side of the fault is in the same direction as the upper side but the dip is 1 in 6. From a point B in the roadway 140 yd West of A a slant road is driven in the seam on a bearing N 50° E and is continued in the same direction and at the same gradient until the seam on the East side of the fault is intersected at a point C. 456 SURVEYING PROBLEMS AND SOLUTIONS Draw to a scale of 1 in. to 100 ft a plan of the roads and fault and mark the point C. State the length of the slant road BC. (M.Q.B./S Ans. 264 yd) 33. The sketch shows a seam of coal which has been subjected to displacement by a trough fault. Calculate the length and gradient of a cross-measures drift to con- nect the seam between A and B from the details shown. (Ans. 587-2 ft; 1 in 2-74) X Y 700' Ex. 8.33 34. A seam dips 1 in 4, 208°30'. Headings at A and B have proved the bearing of the contact line AB to be 075°00*. If the hade of the fault is 30°, what is the true bearing of the fault if (a) it is a downthrow to the South; (b) it is a downthrow to the North. (Ans. 080°42'; 069°18') 35. (a) Define the true and apparent azimuth of a fault. (b) A fault exposed in a certain seam has an azimuth of 086°10', and a hade of 33°. It throws down to the North West. The full dip of the seam is 1 in 6- 5 at 236° 15'. Calculate the true azimuth of the fault. Check by plotting, (c) Two seams, separated by 86 yd of strata, dip at 1 in 13 in a direction S 36° W. They are to be connected by a drift falling at 1 in 5, N 74°30'E. Calculate the plan and slope length of the drift. (N.R.C.T. Ans. (b) 081°12' (c) 330-51 yd 337-07 yd) 36. A mine plan shows an area of 3*6 acres in the form of a square. Measured on the line of full dip underground the length is 632*4 links. Calculate the rate of full dip. ,. - . „ ..„o„wv F (Ans. 1 in 3 or 18°24 ) Bibliography M.H. HADDOCK, The Location of Mineral Fields (Crosby Lockwood). R. McADAM, Colliery Surveying (Oliver & Boyd). FORREST, S.N. Mining Mathematics (Senior Course) (Edward Arnold). METCALFE, J.E., A Mining Engineer's Survey Manual (Electrical Press, London). M.H. HADDOCK, The Basis of Mine Surveying (Chapman & Hall). 9 AREAS 9.1 Areas of Regular Figures The following is a summary of the most important formulae. 9.11 Areas bounded by straight lines Triangle (Fig. 9.1) Fig. 9.1 Triangle (a) (Area) A = half the base x the perpendicular height i.e. A = 1/2 Mi (9.1) (b) A = half the product of any two sides x the sine of the included angle i.e. A = 1/2 ab sin C = 1/2 ac sin B = 1/2 be sin A (9.2) (c) A = yfs(s-a)(s-b)(s-c) (9.3) where s= Wz(a + b+c). Quadrilateral (a) Square, A = side 2 or 1/2 (diagonal 2 ) (9.4) (b) Rectangle, A = length x breadth (9.5) (c) Parallelogram (Fig. 9.2), (i) A = axh (9.6) (ii) A = absina (9-7) / b \ \' f . Fig. 9.3 Trapezium Fig. 9.2 Parallelogram (d) Trapezium (Fig. 9.3) A = half the sum of the parallel sides x the per- pendicular height 458 SURVEYING PROBLEMS AND SOLUTIONS i.e. A = 1/z (a + b)h (e) Irregular quadrilateral (Fig. 9.4) (i) The figure is subdivided into 2 triangles, A = 1/2 {AC x Bb) + uz(AC x Dd) = VzAC(Bb + Dd) a c B (9.8) (9.9) Fig. 9.4 Irregular quadrilateral Fig. 9.5 (ii) A = i/ 2 (/lCxBD)sin0 This formula is obtained as follows, Fig. 9.5: A = uzAX.BX sind + uzBX.CX sin(180 - 0) + 1/2 CX.DX sin 6 + 1/2 DX.AX sin (180 - 0) As sin 6 = sin (180-0), A = 1/2 sin 6 [AX(BX + DX) + CX(BX + DX)] = 1/2 sin 6 [ (AX + CX) (BX + DX)] A = i/zsind AC xBD Regular polygon (Fig. 9.6) (i) A = s/zarxn (9.11) (9.10) /J (ii) As a = 2r tan-^- Area A = i/2rnx2rtan — = nr^tan — 2 2 360 „ «„ = nr 2 tan^— (9.12) 2n (iii) A = 1/2 fl 2 sin 6 x n n 2 360 = 2 R Sin ~n~ (913) Fig. 9.6 Regular polygon (n sides) AREAS 9.12 Areas involving circular corves Circle (i) A = nr z (ii) A = i/4 nd 2 where it ~ 3-1416, 1/477 ~ 0-7854. Sector of a circle. (Fig. 9.7) 459 (9.14) (9.15) (i) A = rrr 2 w 360 (ii) A = ±f* Fig. 9.7 Sector of a circle e (9.16) (9.17) N.B. (ii) is generally the better formula to use as the radian value of is readily available in maths, tables or may be derived from first principles (see Section 2.22). Segment of a circle (Fig. 9.8) Fig. 9.8 Segment of a circle 460 SURVEYING PROBLEMS AND SOLUTIONS (i) A = area of sector — area of triangle i.e. A = i/2r 2 ^ d - i/2r 2 sin0 = 1/2 r 2 (0 - sin 0) (9.18) or A = -nr z i/2r 2 sin(9 360 d = r "ll60- ,/2Sin ^ (9.19) (ii) If chord AC = c and height of arc = h are known, the approximation formulae are: (a) (b) or - Irfffl h 2 2c 3 A ^ W + 4c z h 6c (9.20) (9.21) (9.22) Annulus (flat ring) (Fig. 9.9) A = 7rR 2 -7Tr z = n(R 2 -r 2 ) = 7r(/2-r)(/? + r) (9.23) or A = 1/4 Tr(D-d)(D+d) (9.24) where D = 2R, d = 2r. 9.13 Areas involving non-circular curves Ei/ipse (Fig. 9.10) Fig. 9.9 Annulus 77" A = -ab 4 (9.25) Fig. 9.10 Ellipse where a and ft are major and minor axes. AREAS 461 Parabola (Fig. 9.11) A = -bh 3 (9.26) Fig. 9.11 Parabola 9.14 Surface areas Curved surface of a cylinder (Fig. 9.12) Fig. 9. 12 Curved surface area of a cylinder If the curved surface is laid out flat it will form a rectangle of length 2-77-r = rrD and of height h = height of cylinder. Surface area (S.A.) = lirrh S.A. = vDh (9.27) (9.28) Curved surface of a cone (Fig. 9.13) Fig. 9.13 Curved surface area of a cone 462 SURVEYING PROBLEMS AND SOLUTIONS If the curved surface is laid out flat it will form the sector of a circle of radius /, i.e. the slant height of the cone. By Eq. (9.17), Area of sector = 1/2 r 2 6 Tad = 1/2 rxr0 = i/2r x arc which here = 1/2/ x ttD S.A. = 1/2 nlD i.e. 1/2 x circumference of the base of the cone x slant height. (9.29) Surface area of a sphere (Fig. 9. 14) This is equal to the surface area of a cylinder of diameter D = diameter of the sphere and also of height D. Fig. 9. 14 Surface areas of a sphere A = TTDxD = ttD 2 (9.30) or A = 277Tx2r = Airr 2 (9.31) Surface area of a segment of a sphere In Fig. 9.14, if parallel planes are drawn perpendicular to the axis of the cylinder, the surface area of the segment bounded by these planes will be equal to the surface area of the cylinder bounded by these planes. i.e. S.A. = nDh (9.32) S.A. = 2-rrrh (9.33) where h = the height of the segment. AREAS 463 N.B. The areas of similar figures are proportional to the squares of the corresponding sides (9.34) In Fig. 9. 15, Area A 4BC AB 2 2 Area A AB,C, AB\ BC'' B,C 2 AC ~ACf h 2 A C, C Fig. 9. 15 Areas of similar figures Similarly, in Fig. 9.16, Area of circle 1 = irr 2 = i/4 7rd, Area of circle 2 = 7rr 2 2 = viirtg Area 1 as before Area 2 rl dl (9.35) (9.36) Fig. 9.16 Comparison of areas of circles Example 9.1. The three sides of a triangular field are 663*75 links, 632*2 links and 654*05 links. Calculate its area in acres. a = 663*75 s -a = 311*25 b = 632*20 s - b = 342*80 c = 654*05 s-c = 320*95 2U950-00 = 975*00 s = 975*00 464 SURVEYING PROBLEMS AND SOLUTIONS By Eq. (11.3), Area = V s(s - a)(s - b)(s - c) = V975-00 x 311-25 x 342-80 x 320-95 = 182724 link 2 = 1-827 acres Example 9.2 A parallelogram has sides 147-2 and 135-7 ft. If the acute angle between the sides is 34°32' calculate its area in square yards. By Eq. (11.7), Area = ab sin a = 147-2 x 135-7 sin 34°32' = 11323 ft 2 = 1258 yd 2 Example 9.3 The area of a trapezium is 900 ft . If the perpendicular distance between the two parallel sides is 38 ft find the length of the parallel sides if their difference is 5 ft. ByEq.(9.8), 3g A = \ x + (x - 5) { — = 900 1800 .-. 2x - 5 = — — 38 47-368 + 5 52-368 x = 2 2 = 26-184 Lengths of the parallel sides are 26-184 and 21-184. Example 9.4 ABC is a triangular plot of ground in which AB mea- sures 600 ft, (182-88 m). If angle C = 73° and angle B = 68° find the area in acres. In triangle ABC, AB sin A 600 sin {180 - (73 + 68)} BC = sin C sin 73 600 sin 39° sin 73° = 394-85 ft (120-345 m) AREAS 465 ByEq.(9.2), Area of triangle ^BCABsinB 2 -x 394-85x600 sin 68 2 109829 ft 2 109829 9 x 4840 2*521 acres. (10 203 m 2 ) acres Example 9.5 Calculate the area of the underground roadway from the measurements given. Assume the arch to be circular. In the segment (Fig. 9. 18) h = 7'- 10"- 5' = 2'- 10" c = 12'- 0" "-tf*© 1 = r 2 - 2rh + h z + -c z h 2 + ±c z 2-833 2 + ^ 4 n 2h 2 x 2-833 = 7-769 ft Sin- = 2 c 12 2r 2 x 7-769 = 0-77 230 e 2 ~ 50°33'40" 6 = 101°07'20' /^ ^X % " o "Ks U) " " 12-0" " Fig. 9. 17 .» ,n ^" ^^^ ^ yr ^' ^r ^^ /» N. Z''' X. c >v 2 a r-/> >^ rX. 2 Fig. 9.18 Area of segment = - r\6 - sin 0) 7-769 2 (1-76492-0-98122) = ■ 30-179 x 0-78370 = 23 ' 651 ft2 Area of rectangle = 5x12 = 60 ft 2 Total area = 83-651 ft 2 466 SURVEYING PROBLEMS AND SOLUTIONS Check By Eq.(9.20), Area of segment ByEq.(9.22), Area of segment 4 //5x2-833\ 2 . - fx 2-833 VV~I j +6 ' ^ 3-776 V3-135 + 36 ~ 3-776 x 6-294 ~ 23-77 ft 2 3 x 2-833 3 + 4 x 144 x 2-833 6x12 ~ 23-6 lft 2 Example 9.6 A square of 6 ft sides is to be subdivided into three equal parts by two straight lines parallel to the diagonal. Calculate the perpendicular distance between the parallel lines. Triangle ABC = - area of square Triangle BEF = - area of square Length of diagonal AC = 6yf2 Length BJ = 3/2 = 4-242 Area of A BFE 2 Area of A ABC 3 BK 2 2 BJ 2 3 BK* = 4-242' 1-5 BK = 4-242 x y/VS VS = 3-464 ft KJ = BJ-BK = 4-242-3-464 Width apart of parallel lines Fig. 9.19 0-778 1-556 ft AREAS 467 Example 9.7 The side of a square paddock is 65 yd (59*44 m). At a point in one side 19V2yd (17*83 m) from one corner a horse is tether- ed by a rope 39 yd (35*66 m) long. What area of grazing does the horse occupy? F B Fig. 9.20 The area occupied consists of a right-angled triangle AEG + ^ sector EFG. 19y 2 In triangle AEG cos £ = — - =0*5 £ = 60° ByEq.(9.2), Area of triangle AEG = \ x 19 x / 2 x 39 sin 60 = 329*31 yd 2 (275*34 m 2 ) ByEq.(9.17), Area of sector 1 x 39 2 x 120° 'red Total area = 760*5x2*094 = 1592*49 yd 2 (1331*52 m 2 ) = 1921*80 yd 2 (1606*86 m 2 ) of grazing Example 9.8 In order to reduce the amount of subsidence from the workings of a seam the amount of extraction is limited to 25% by driv- ing 12ft wide roadways. What must be the size of the square pillars left to fulfil this condition? 468 SURVEYING PROBLEMS AND SOLUTIONS JJI Pillar TF 6'^- -fir ii c r Fig. 9.21 N.B. The areas of similar figures are proportional to the squares of the relative dimensions, 100 i.e. (* + 12)' 75 x 2 i.e. 100x 2 = 75(x 2 + 24* + 144) = 75x 2 + 1800*+ 10800 25x 2 - 1800* - 10800 = x 2 - 72x- 432 = By the formula for solving quadratic equations. -b±y/b 2 - 4ac 2a 72 , c = -432, a = 1, 72 +V5184 + 1728 A." 2 72 ±V6912 2 = 36 ±41-57 = 77-57 ft Alternative Solution Size of square representing 100% Size of square representing 75% Difference in size Actual difference in size Vioo = 10 (Am V75 = 8-66 (EF) = 1-34 = 12 ft AREAS 469 If x is the size of the pillar, 1'34 : 12 12 x 8-66 8-66 : x 1-34 77-57 ft Example 9.9 A circular shaft is found to be 4 ft out of vertical at the bottom. If the diameter of the shaft is 20 ft, find the area of the crescent-shaped portion at the bottom of the shaft which is outside the circumference from the true centre at the surface. Fig. 9.22 Area of crescent-shaped portion = Area of circle - 2 (area of seg- ment AXB). Area of circle = TTr 2 = 3-1416x100 = 314-16 ft 2 Area of sector A XBQ = -r z 6 1 2 -i oy -i 2 6/2 = cos — = cos — OA 10 2 e Area of sector AX BO, Area of triangle OAB Area of segment AXB Area of crescent 78°32' 157°04' 1100x2-74133 137-07 Il00sinl57°04' = 19-48 = 117-59 314-16 - 2x117-59= 78-98 ft 2 470 SURVEYING PROBLEMS AND SOLUTIONS Example 9.10 In a quadrilateral ABCD.A = 55°10', B = 78°30' C = 136°20' . (A and C are diagonally opposite each other) AB = 620 links, DC = 284 links. (a) Plot the figure to scale and from scaled values obtain the area in acres. (b) Calculate the area in acres. ISO-^' K» 78*30')«= 46*20' Construction (N.B. D = 360 - (55° 10' + 78°30' + 136°20') = 90°) Draw a line parallel to AD 284 links away. This will cut AE at C. From scaling, AC = 637 links Dd = 256 links Bb = 296 links 552 Area = \ (637 x 552) = 175810 sq links = 1*758 1 acres Calculation In triangle AEB, AE = AB sin B 620 sin78°30' sinfi sin46°20' = 839-89 Iks Area of triangle AEB = ^AE.AB sin A = 1839-89 x 620 sin 55°10' = 213713 sq Iks In triangle DEC, DE = DC cot 46°20' = 284cot46°20' = 271-08 Iks Area of triangle DEC = AREAS + DE.DC 1 271-08 x 284 2 Area of triangle ABCD = 213713 - 38493 471 = 38 493 sq Iks = 175 220 sq Iks = 1-7522 acres 9.2 Areas of Irregular Figures In many cases an irregular figure can conveniently be divided into a series of regular geometrical figures, the total area being the sum of the separate parts. If the boundary of the figure is very irregular the following methods may be employed. 9.21 Equalisation of the boundary to give straight lines (Fig. 9.24) Fig. 9.24 Equalisation of an irregular boundary 472 SURVEYING PROBLEMS AND SOLUTIONS The irregular boundary A B C D E F is to be equalised by a line from a point on YY to F. Construction (1) Join A to C; draw a line Bb parallel to AC cutting YY at b. Triangle AbC is then equal to triangle ABC. (triangles standing on the same base and between the same parallels are equal in area). (2) Repeat this procedure. Join bD, the parallel through C to give c on YY. (3) This process is now repeated as shown until the final line eF equalises the boundary so that area eABX = area XFEDC N.B. (a) This process may be used to reduce a polygon to a triangle of equal area, (b) With practice there is little need to draw the construction lines but merely to record the position of the points b,c,d, e, etc. on line YY. Reduction of a polygon (Fig. 9.25) b A Ed Fig. 9.25 Reduction of a polygon to a triangle of equal area Area of triangle bCd = Area of polygon ABCDE. (4) Where the boundary strips are more tortuous the following methods may be adopted. 9.22 The mean ordinate rule (Fig. 9.26) The figure is divided into a number of strips of equal width and the lengths of the ordinates o,, o 2 , o 3 etc. measured. (N.B. If the beginning or end of the figure is a point the ordinate is included as q = zero.) AREAS 473 Fig. 9.26 The mean ordinate rule The area is then calculated as (Ol+0 2 +0 3 + 4 + ...O n ) f\ = -^— — ^— — — -^^^— — — — — — — — - n where n = number of ordinates 2 ordinates x W x(n-l)w or A = (9.37) (9.38) where W = 2 w. This method is not very accurate as it implies that the average ordinate is multiplied by the total width W. 9.23 The mid-ordinate rule (Fig. 9.27) Here the figure is similarly divided into equal strips but these are then sub-divided, each strip having a mid-ordinate, aa, bb, cc etc. The average value of these mid- ordinates is then multiplied by the total width W. b . d a - ^ j a^««. ' S^i 1 6" C w i.e. Area = mW where m = the mean of the mean ordinates. (9.39) The only advantage of this meth- Fig- 9.27 The mid-ordinate rule od is that the number of scaled values is reduced. 9.24 The trapezoidal rule (Fig. 9.28) 01 02 . w » » 03 Fig. 9.28 The trapezoidal rule 474 SURVEYING PROBLEMS AND SOLUTIONS This is a mote accurate method which assumes that the boundary between the extremities of the ordinates are straight lines. The area of the first trapezium A y «<m (o, + o 2 ) The area of the second trapezium A 2 w, 2 "<o 2 + o 3 ) The area of the last trapezium 4»-i = ~7T~ (°n-i + °n) If w, =w z = w 3 = w n = w; w then the total area = — [o, + 2o 2 + 2o 3 +...2o n _, + <^] (9.40) 9.25 Simpsorfs rule (Fig. 9.29) This assumes that the boundaries are curved lines and are considered as portions of parabolic arcs of the form y = ax 2 + bx + c. The area of the figure Aab t cCB is made up of two parts, the trapez- ium AabcC + the curved portion above the line abc. Fig. 9.29 Simpsorfs rule The area of ab.cb = - of the parallelogram aa. b. c. cb. 1 O 111 4vv [<*> - {(°y + °3>] [1 -I 4\V r t I 2(°i + ° 3 )J + ~ [°2 ~ 2<°i + °3)J W, = 3"t°i + 4o z + ° 3 1 (9.41) N.B. This is of the same form as the prismoidal formula with the linear values of the ordinates replacing the cross-sectional areas. AREAS 475 If the figure is divided into an even number of parts giving an odd number of ordinates, the total area of the figure is given as w A = [(o, + 4o 2 + o 3 ) + (03 + 4o 4 + 05)+... o n _ 2 + 4o n _, + qj w -■ .-. A = [q + 4o 2 + 2q, + 4o 4 + 2o 5 +... 2o n _ 2 + 4o n _, + qj (9.42) The rule therefore states that: a if the figure is divided into an even number of divisions, the tot- al area is equal to one third of the width between the ordinates mult- iplied by the sum of the first and last ordinate + twice the sum of the remaining odd ordinates + four times the sum of the even ordinates" This rule is more accurate than the others for most irregular areas and volumes met with in surveying. Example 9.11 A plot of land has two straight boundaries AB and BC and the third boundary is irregular. The dimensions in feet are AB « 720, BC = 650 and the straight line CA = 828. Offsets from CA on the side away from B are 0, 16, 25, 9, feet at chainages 0, 186, 402, 652, and 828 respectively from A. (a) Describe briefly three methods of obtaining the area of such a plot. (b) Obtain, by any method, the area of the above plot in acres, ex- pressing the result to two places of decimals. (R.I.C.S.) The area of the figure can be found by (1) the use of a planimeter; (2) the equalisation of the irregular boundary to form a straight line and thus a triangle of equal area (3) the solution of the triangle ABC + the area of the irregular boundary above the line by one of the ordinate solutions. Fig. 9.30 476 SURVEYING PROBLEMS AND SOLUTIONS By (2), Area = 1(754 x 625) = 235 625 ft 2 = 5-41 acres By (3), Area of triangle ABC = y/s(s - a)(s - b)(s - c) (Eq. 9.3) where s = -(a + b + c) i.e. a = 650 s-a = 449 b = 828 s-b = 271 c = 720 s-c = 379 2)2198 s = 1099 Check s = 1099 Area = V 109 ^ x 449 x 271 x 379 = 225126 ft 2 Area of the irregular boundary (a) By the mean ordinate rule, Eq. (9.38), + 16 + 25 + 9 + Area = x 828 = 8280 ft 2 Total Area 233406 ft 2 = 5*35 acres (b) By the trapezoidal rule, Eq. (9.40), Area = 1[186(0 + 16) + (402 - 186) (16 + 25) + (652 - 402) (25 + 9) + (828 - 652) (9 + 0)] = 1[186 x 16 + 216 x 41 + 250 x 34 + 176 x 9] = \ [2976 + 8856 + 8500 + 1584] = 10958 ft 2 Total Area 236 084 ft 2 = 5*43 acres N.B. As the distance apart of the offsets is irregular neither the full Trapezoidal Rule nor Simpson's Rule are applicable. AREAS 477 9.26 The planimeter This is a mechanical integrator used for measuring the area of ir- regular figures. It consists essentially of two bars OA and AB, with fixed as a fulcrum and A forming a freely moving joint between the bars. Thus A is allowed to rotate along the circumference of a circle of radius OA whilst B can move in any direction with a limiting circle OB. Theory of the Planimeter (Fig. 9.31) Let the joint at A move to A y and the tracing point B move first to B, and then through a small angle 5a to B 2 . If the whole motion is very small, the area traced out by the tracing bar AB is ABB.B^ i.e. ABx 8h + ±AB z 8a or 8A = I8h + jl 2 8a (9.43) where 8A = the small increment of area, / = the length of the tracing bar AB, 8h = the perpendicular height between the parallel lines AB and j4,B„ 5a = the small angle of rot- ation. Fig. 9.31 Theory of the planimeter A small wheel is now introduced at W on the tracing bar which will rotate, when moved at right-angles to the bar AB, and slide when moved in a direction parallel to its axis, i.e. the bar AB. Let the length AW = k.AB. Then the recorded value on the wheel will be 8W = 8h + kAB8a i-e. 8h = 8w _ kAB8a = 8w - kl8a which when substituted in Eq. (9.43) gives 54 = l(8w-U8a) + U 2 8a = I8w+ Z 2 (i -k)8a (9.44) 478 SURVEYING PROBLEMS AND SOLUTIONS To obtain the total area with respect to the recorded value on the wheel and the total rotation of the arm, by integrating, A = lw + f(i-kja (9.45) where A = the area traced by the bar, w = the total displacement recorded on the wheel, a = the total angle of rotation of the bar. Two cases are now considered: (1) When the fulcrum (0) is outside the figure being traced. (2) When the fulcrum (0) is inside the figure being traced. (1) When the fulcrum is outside the figure (Fig. 9.32) Commencing at a the joint is at A. Moving to the right, the line abed is traced by the pointer whilst the bar traces out the positive area 04,) abcdDCBA. Moving to the left, the line defa is traced out by the pointer whilst the negative area (4 2 ) defaAFED is traced out by the bar. The difference between these two areas is the area of the figure abedefa, i.e. Fig. 9.32 Planimeter fulcrum outside the figure A = 4, - A z = lw + i ! (i-*)a but a = A = lw (Eq. 9.45) (9.46) N.B. The joint has moved along the arc A BCD to the right, then along DEFA to the left. In measuring such an area the following procedure should be follow- ed: (1) With the pole and tracing arms approximately at right-angles, place the tracing point in the centre of the area to be measured. (2) Approximately circumscribe the area, to judge the size of the area compared with the capacity of the instrument. If not poss- ible the pole should be placed elsewhere, or if the area is too AREAS 479 large it can be divided into sections, each being measured sep- arately. (3) Note the position on the figure where the drum does not record - this is a good starting point (A). (4) Record the reading of the vernier whilst the pointer is at A. (5) Circumscribe the area carefully in a clockwise direction and again read the vernier on returning to A. (6) The difference between the first and second reading will be the required area. (This process should be repeated for ac- curate results). (7) Some instruments have a variable scale on the tracing arm to give conversion for scale factors. (2) When the fulcrum is inside the figure (Fig. 9.33) Fig. 9.33 Planimeter fulcrum inside the figure In this case the bar traces out the figure (A T ) abcdefgha - the area of the circle (4) A BCDEFGHA; it has rotated through a full circle, i.e. a = 2 77. A T -A C = l W + l 2 (l- k\2TT .: A T = Iw + f (j- k\2n+ A c = l w + l 2 (l- }S 2n + -nb 2 (where b = 04) = Iw + n{b 2 +fa-2k)] (9.47) 480 SURVEYING PROBLEMS AND SOLUTIONS This is explained as follows Fig. 9.34: If the pointer P were to rotate without the wheel W moving, the angle OWP would be 90° The figure thus described is known as the zero circle of radius r; Fig. 9.34 Theory of the zero circle or r 2 = b 2 -(klf = OW z +(l-klf = b z - (ktf + I 2 - 2kl 2 + (klf = b 2 + I 2 (1 -2k) In Eq. (9.47), An = lw + nr = Iw + the area of the zero circle The value of the zero circle is quoted by the manufacturer. (9.48) (9.49) N.B. (1) A T - A c = lw. If A c > A T then lw will be negative, i.e. the second reading will be less than the first, the wheel having a resultant negative recording. (2) The area of the zero circle is converted by the manufactur- er into revolutions on the measuring wheel and this con- stant is normally added to the recorded number of revolu- Example 9.12 /irp ^> ™-o 1st reading 3-597 2nd reading 12-642 Difference 9-045 Constant Total value 1st reading 23-515 32-560 Arp < Aq 6-424 2nd reading 3-165 Difference Constant -3-259 23-515 Total value 20-256 AREAS 481 9.3 Plan Areas 9.31 Units of area lsq foot (ft 2 ) = 144 sq inches (in 2 ) lsqyard(yd 2 ) = 9 ft 2 1 acre = 4 roods = 10 sq chains = 100 000 sq links = 4840 yd 2 = 43 560 ft 2 1 sq mile = 640 acres Conversion factors lin 2 = 6-4516 cm 2 1 cm 2 = 0-155000 in 2 lft 2 = 0-092 903 m 2 lyd 2 = 0-836 127 m 2 1 m 2 = 1-19599 yd 2 1 sq chain = 404-686 m 2 1 rood = 1011-71 m 2 1 acre = 4046-86 m 2 1 km 2 = 247-105 acres 0-404686 hectare (ha) 1 ha = 2-47105 acres lsq mile = 2-58999 km 2 = 258-999 ha N.B. The hectare is not an S.I. unit. The British units of land measurement are the Imperial Acre and the Rood (the pole or perch is no longer valid). The fractional part of an acre is generally expressed as a decimal although the rood is still valid. Thus 56-342 acres becomes 56-342 acres _4_ 56 acres 1-368 roods The use of the Gunter chain has been perpetuated largely because of the relationship between the acre and the square chain. Thus 240362 sq links = 24-036 2 sq chains = 2-40362 acres The basic unit of area in the proposed International System is the square metre (m 2 ). 482 SURVEYING PROBLEMS AND SOLUTIONS 9.32 Conversion of planimetric area in square inches into acres Let the scale of the plan be — . x i.e. lin. = xin. lsqin. = x 2 sqin. v-2 12 x 12 x 9 x 4840 acres Example 9.13 Find the conversion factors for the following scales, (a) 1/2500 (b) 6 in. to 1 mile, (c) 2 chains to 1 inch. 2500 2 (a) lin 2 = — — — — - — — — = 0-995 acres 12 x 12 x 9 x 4840 9 (4026-6 m 2 = 0-4026 ha) (b) 6in.tol.nile Q^ . 1760x36 1Ar ^ n i.e. lin. = = 10 560 in. 6 10560 2 = 17-778 acres 144x43560 (71945 m 2 = 7-1945 ha) Alternatively, 6 in. = 1 mile 36 in =1 mile 2 = 640 acres . 2 640 1 in 2 = — — = 17-778 acres 36 (c) 2 chains to 1 inch. lin. = 200 links 1 in 2 = 40 000 sq links = 0-4 acres (1618-7 m 2 = 0-16187 ha) Alternatively, 2 chains to 1 inch (1/1584) lin. = 2x66 ft = 132x12 = 1584ft 1 • 2 1584 2 1 in* = . = 0-4 acre 144x43560 9.33 Calculation of area from co-ordinates Method 1. By the use of an enclosing rectangle (Fig. 9.35) AREAS 483 B(x&)W Ofx»*J 1 E(x B *) Fig. 9.35 Calculation of area by enclosing rectangle The area of the figure ABCDE = the area of the rectangle VWXYZ -{MVB + BWXC + ACXD + ADYE + MEZ I This is the easiest method to understand and remember but is laborious in its application. Method 2. By formulae using the total co-ordinates Applying the co-ordinates to the above system we have: Area of rectangle VWXYZ = (x A -xJ(y 2 -y g ) of triangle AVB = l(x 2 -x,)(y 2 -y 1 ) of trapezium BWXC = - (y 2 - y 3) { (x 4 - x-j) + (x 4 - x 3 ) \ of triangle CXD = I(*-y 4 )(*-*) DYE = l(y 4 -*)<*,-*) AEZ = £<*-*)(*-*,) i.e. A = (x 4 )2 - x 4 y 8 - x,^ + x,y 5 ) - i[ x^ - x^, - x,)*, + x,y, + 2x 4 y2 - 2x 4 y 3 - x z y 2 + x 2 y 3 -x a y 2 + x 3 y 3 + x 4 y 3 - x 4 y A -x 3 y 3 + x 3 y 4 + ^ y 4 - *4 y 5 " x 5 y 4 + x s y 5 + Xgy, - x^ 6 -*iy,+*,yj •*• 4 = i[y,(x 2 -x 6 ) + y 2 (x3-x ) ) + y 3 (x 4 -x 2 ) + y 4 (x B -x 3 ) + y 5 (x 1 -x 4 )] 484 SURVEYING PROBLEMS AND SOLUTIONS This may be summarised as A = £2&<* f ^-i) (9.50) i.e. Area = half the sum of the product of the total latitude of each station x the difference between the total departures of the preceding and following stations. This calculation is best carried out by a tabular system. Example 9.14 The co-ordinates of the corners of a polygonal area of ground are taken in order, as follows, in feet: A (0, 0); B (200, -160); C (630, -205); D (1000, 70); E (720, 400); F (310, 540); G (-95, 135), returning to A. Calculate the area in acres. Calculate also the co-ordinates of the far end of a straight fence from A which cuts the area in half. To calculate the area of the figure ABCDEFG the co-ordinates are tabulated as follows: (a) (b) (c) (d) (c)-(d) (a)x{(c)-(d)} Station T.Lat. T.Dep. Preceding Following + A B C D E F G -160 -205 70 400 540 135 200 630 1000 720 310 -95 Dep. 200 630 1000 720 310 -95 Dep. -95 200 630 1000 720 310 295 630 800 90 -690 -815 -310 6300 100800 164000 276 000 440 100 41850 6300 1022750 6 300 2 ) 1016 450 508 225 ft 2 .-. Total Area = 508225 ft 2 (47 215-63 m 2 ) = 11-667 acres (4-72156 ha) From a visual inspection it is apparent that the bisector of the area AX will cut the line ED. The area of the figure A BCD can be found by using the above figures. AREAS 485 A B C D 200 1000 -800 160 200 630 630 205 630 1000 200 800 70 1000 630 -630 100800 164000 44100 2)308900 154450 ft 2 J^310,540) / ^^^«f(720,4OO) 6 <f <-95,135)\ ,''''' * {1O00,70) A (0,0) B <S (200,-160) (630,-205) Fig. 9.36 The area of the triangle AXD must equal ^(508 225) - 154 450 = 99 662-5 ft 2 Also area = ^AD.DX sin D DX = 2x99662-5 AD sin D To find length AD and angle D. Bearing DA = tan -1 -1000/ -70 = S 85°59'50" W = 265°59'50" Length DA = 1000/sin85°59'50" = 1002-45 ft Bearing DE = tan" 1 - 280/330 = N 40°18'50" W = 319°41'10" Angle 2x99662-5 D DX 53°41'20' 1002-45 sin 53°4l'20 ■„ = 246-76 ft To find the co-ordinates of X. (N 40°18'50" W 246*76 ft). E x = 840-35 ft sin bearing 0*64697 cos bearing 0*762 51 E D = 1000*00 AE = -159*65 AN = +188*16 N n = 70-00 N, 258 -16 ft 486 SURVEYING PROBLEMS AND SOLUTIONS Ch eck on Area A 200 840-35 -640-35 B -160 200 630 630 100800 C -205 630 1000 200 800 164000 D 70 1000 840-35 630 210-35 14724-5 X 258-16 840-35 1000 - 1000 258 160 522960 14724-5 508 235-5 ft 2 Method 3. By areas related to one of the co-ordinate axes (Fig. 9.37) Fig. 9.37 Areas related to one axis Area ABCDE = (bBCc + cCDd + dDEe) - (bBAa + aAEe) Using the co-ordinates designated, trapeziums ( bBAa = '-(x 2+ x,)(y 2 -y,) aAEe = Ux, +*,)(y, -y 5 ) bBCc = |(x 2+ x 3 )(y 2 -y 3 )' cCDd = ^(x 3 +x A )(y 3 -y A ) , -{ dDEe = ^(* 4 +*5)(y 4 -y 5 ) t i- e. j[ (*2M> - x 2 y 3 + x 3 y 2 - x^) + (x^ - x 3 y A + x A y 3 - x+y A ) + (x A y A - xjs + %y A - a^y 3 ) - (x,y 2 - x,y, + x,^ - x,y, ) -O^-^ys+^y,-*.^)] (9.51) Uyi(x 2 -xJ + y 2 (Xa -x t ) + 3a (x A -x 2 ) + y^-xj + ^(x, -x A )] AREAS 487 as before Area ^y n (x n+1 -^_,) (Eq.9.50) Method 4. Area by 'Latitudes* and 'Longitudes* (Fig. 9.38) Here Latitude is defined as the partial latitude of a line Longitude is defined as the distance from the y axis to the centre of the line. Fig. 9.38 Calculation of area by latitudes and longitudes From Eq. (9.51), A = \l\{x 2 + x 3 )iy z -y 3 )^{x 3 + x A ){y 3 -y A ) + {x A +x^(y A -y z )\ - {(^ + x,)(y 2 -y l ) + (x, + x 5 )(y 1 -y 5 )}] = l[(x, +x z )(y, -y 2 ) + (x 2 + x 3 )(y 2 -y 3 ) + (x 3 + x A )(y 3 -y 4 ) + (* 4 + * 5 ) (y 4 - y 5 ) + (* 5 + *i> (y 8 " ?f > 1 = -;s(x n + v ,)(yn-yn + ,) ( 952 > where ^(x n + x n+1 ) = the longitude of a line. Ok - yn+i) = the partial latitude of a line, i.e. the latitude. N.B. (1) It is preferable to use double longitudes and thus produce double areas, the total sum being finally divided by 2 (2) This method is adaptable for tabulation using total depar- tures and partial latitudes or vice versa. 488 SURVEYING PROBLEMS AND SOLUTIONS Example 9.15 From the previous Example 9.14 the following table is compiled. Stn. T. Dep. P. Lat. Sum of Double Area Adj. T. Dep. + - A -160 200 32000 B 200 -45 830 37 350 C 630 + 275 1630 448 250 D 1000 + 330 1720 567600 E 720 + 140 1030 144 200 F 310 -405 215 87075 G -95 -135 -95 12825 A 1172875 156 425 156425 2)1016 450 508 225 ft 2 9.34 Machine calculations with checks Using Eq. (9.52), (1) (2) (3) (4) (5) = (3)x(4) (6) (7) (8) = (6)x(7) y X dy Ix dx Sy A -160 200 - 32 000 200 -160 - 32 000 B -160 200 - 45 830 - 37 350 430 -365 -156 950 C -205 630 + 275 1630 448 250 370 -135 -49 950 D 70 1000 + 330 1720 567 600 -280 470 -131600 E 400 720 + 140 1030 144 200 -410 940 - 385 400 F 540 310 -405 215 -87 075 -405 675 -273 375 G 135 -95 + 2860 -135 -95 + 12825 95 135 + 2220 + 12 825 + 1145 + 745 + 5625 + 1172875 + 1095 + 12 825 -365 + 780 -95 + 2765 -745 -95 + 5530 -156 425 -1095 -660 - 1 029 275 + 1 016 450 1560 1016 450 (X2) (X2) 1560 5530 508 225 fl 2 508 225ft 2 AREAS 489 N.B. (1) From the total co-ordinates in columns 1 and 2, the sum and difference between adjacent stations are derived. e.g. dy AB = (-160-0) = -160 Zy AB = -160+0 = -160 dy B c= (-205+160) = - 45 Sy sc = -160-205 = -365 (2) All the arithmetical checks should be carried out to prove the insertion of the correct values. (a) 2 x the sum of the total latitudes = Sy (col. 7). (b) 2 x the sum of the total departures = Sx (col. 4). (c) The Algebraic sum of columns 3 and 6 should equal zero. (3) The product of columns 3 and 4 gives column 5, i.e. the double area: Also the product of columns 6 and 7 gives column 8, i.e. the double area: Columns 5 and 8 when totalled should check. Alternative method From the previous work it is seen that the area is equal to one- half of the sum of the products obtained by multiplying the ordinate (latitude) of each point by the difference between the abscissae (departure) of the following and preceding points. i- e - A = ^iW ~ *>y, + x 3 y 2 - x, y 2 + x A y 3 - x z y, 2^3 + x 5 y A - x& A + x, yg - x 4 y 5 ) (9. 53) This may be written as, A = L 1 v ^r* 2 - :i (9.54) This is interpreted as 'the area equals one half of the sum of the pro- ducts of the co-ordinates joined by solid lines minus one half of the sum of the products of the co-ordinates joined by dotted lines, This method has more multiplications but only one subtraction. Using the previous example; A = a TO.^ 200^ 630^ 1000^ 720 310^ _ 95"] 2 U^-160' > >-205' > ^ 70 ""400^^540^^135 J = ^[-182700 -833750] = -[ 1 016 450 ] (negative sign neglected) 2 = 508 225 ft 490 SURVEYING PROBLEMS AND SOLUTIONS 9.4 Subdivision of Areas* 9.41 The subdivision of an area into specified parts from a point on the boundary (Fig. 9.39) B Fig. 9.39 Subdivision of an area from a point on the boundary Consider the area ABCDE to be equally divided by a line start- ing from X on the line ED. (1) Plot the co-ordinates to scale. (2) By inspection or trial and error decide on the approximate line of subdivision XY. (3) Select a station nearest to the line XY, i.e. A or B. (4) Calculate the total area ABCDE. (5) Calculate the area AXE. (6) Calculate the area AXY = \ABCDE - AXE. (7) Calculate the length and bearing of AX. (8) Calculate the bearing of line ED. (9) Calculate the length A Y in triangle AYX. N.B. Area of triangle AYX = -AX.XY sin X. As the area, AX and X are known, AY is calculated: AY Area of triangle AYX ,g r« -AX sin X 2 (10) Calculate the co-ordinates of Y. * For a complete analysis of this the reader is advised to consult The Basis of Mine Surveying by M.H. Haddock. AREAS 491 9.42 The subdivision of an area by a line of known bearing (Fig. 9.40) Fig. 9.40 Subdivision of an area by a line of known bearing Construction Given the area ABCDE and the bearing of the line of sub-division XY, then EF on the given bearing and XY will be parallel to this, a perpendicular distance d away. Draw FG perpendicular to XY. HX perpendicular to EF. The area AYXE = ^area ABCDE = AAFE + AFYG+FGXH + AHXE. (1) From the co-ordinates the length and bearing of AE can be calculated. (2) In the triangle AFE the area can be obtained by first solving for the length EF. (3) The area of the figure FY GX EH can thus be obtained in terms of d, i.e. triangle FYG = -d cot a rectangle FGXH = d(EF - HE) = d(EF- dcotB) triangle HXE = ^d 2 cot)8 492 SURVEYING PROBLEMS AND SOLUTIONS .'. dEF + irf 2 (cota-cot)3) = Area of AYXE - Area of AAFE. (9.56) This is a quadratic in d as the angles a and B are found from the bearings. From the value of d the co-ordinates of F and E can be obtained. 9.43 The subdivision of an area by a line through a known point inside the figure (Fig. 9.41) Fig. 9.41 Subdivision of an area by a line through a known point inside the figure Construction Given the area ABC D E and the co-ordinates of the fixed point H, join EH and produce to cut AB in G. Assume the dividing line XY is rotated <x° about H. From the co-ordinates: (1) Calculate the length and bearing EH. (2) In the triangle AGE calculate the length EG (this gives the length AG) and the area. (3) The required area AYXE = A AGE - A YGff + AE//X = A 4GE- area (A) To find the missing area, A = \EH.HX sin a +±GH.HY sin a ! _ EH sin <t> sin a i GH sin 6 sin a = |E// x , x + ^Gtf x 2 sin(a+0) z sin(a+0) AREAS 493 EH Z GH' cot a + cot <f> cot a + cot 6 .] 24 = EH' GH' • + cot a + cot (9.57) cot a + cot <fi As the lengths EH and GH are known, and <f> and are obtain- able from the bearings, this is a quadratic equation in cot a, from which a may be found, and thus the co-ordinates of X and Y. These problems are best treated from first principles based on the foregoing basic ideas. Example 9.16. In a quadrilateral A BCD, the co-ordinates of the points, in metres, are as follows: Point E N A B -893-8 C +634-8 -728-8 D +1068-4 +699-3 Find the area of the figure by calculation. If E is the mid-point of AB, find, graphically or by calculation, the co-ordinates of a point F, on the line CD, such that the area AEFD equals the area EBCF. N.B. Co-ordinates of E = -(A + B) i.e. 0, ^x -893-8 = 0, 446-9 '(1068-4, 699-3) 26° 06' [892-5,119-8] (634-8,-728-8) Fig. 9.42 494 SURVEYING PROBLEMS AND SOLUTIONS x y dx iy dy ix A -893-8 - -893«8 104 742-00 B -893-8 634-8-1622-6-1030 026-48 165-0 634-8 2432339-92 C 634-8 -728-8 433-6 -29-5 -12 791-20 1428-1 1703-2 D 1068-4 +699-3 - 1068-4 699-3 -747 132-12 -699-3 1068-4 -747 132-12 1703-2 699-3 1068-4 -699-3 1593-1 3406-4 2537081-92 -1622-6 -1068-4 -2545-9 -1789 949-80 -1593-1 -747 132-12 1703-2 -923-3 - 1846-6 - 1 789 949-80 3406-4 +1789 949-80 3406-4-1846-6 Area = 894 974-9 it 2 Referring to Fig. 9.42, . 1068-4 - Bearing ED = tan" 1 —— -— - = tan -1 0-93212 = N 42°59'20"E & 699-3 + 446-9 Length ED = 1068-4 sin 42°59'20" = 1566-9 In triangle ADE, Area = -AE.ED sin 42°59'20" = i x 446-9 x 1566-9 sin42°59'20" = 238735-4 m 2 -•- Area triangle EDF = 894 974'9 _ 238 735-4 sq ft = 208752 m 2 , 634-8 - 1068-4 Bearing DP = Bearing DC = tan ,728-8 -699-3 = tan"' 0-30362 = S le^^O" W Angle ADF = 42°59'20" - 16°53'20" = 26 o 06 , Using Eq. (9.55), Area A EDF _ 2x208752 DF = ^-ED sin EDF = 1566 ' 9 sin 26 ° 06 ' = — " 2 To obtain the co-ordinates of F, Line DF S 16°53'20" W Length 605-66 m P. Dep. 605-66 sin 16°53'20" = -175-9 P. Lat. 605-66 cos 16°53'20" = -579-5 T. Dep. of F 1068-4 - 175-9 = 892-5 m T. Lat . of F 699-3-579-5 = 119-8 m Example 9.17 With the previous co-ordinate values let the bearing of the divid- ing line be N 57°35'10" E. AREAS 495 Construction Draw line AG on this bearing and EF parallel to this a perpend- icular distance d away. . 1068-4 Bearing AD = tan 599.3 = tan" 1 1-52781 = N 56°47'40" E Length AD = 1068-4 cosec56°47 '40 = 1276-9 m Fig. 9.43 In triangle ADG, A = 57°35'10" -56°47'40" = 0°47'30" D = 56°47'40" - 16°53'20" = 39°54'20" G = 180 -(57 o 35'10" - 16°53'20") = 139°18'10" 180°00 , 00" AD sin D 1276-9 sin 39°54'20" AG sin 6* sin 139°18'10" = 1256-3 m Area triangle ADG = ^AD.AG sin A = jx 1276-9 x 1256-3 sin0°47'30" = 11084-8 m 2 Area AGFE = ± Area ABCD -A ADG 2 = 447487-5-11084-8 = 436402-7 m a In figure AGFE, Area = AA7E + AHFJ + AFHG 1 J2 ' j2 = ± d z cot £ + d( AG -d cot a) + -d* cot a = 1256-3 d + \ d\cot j8 - cot a) where a = 57°35'10" - 16°53'20" = 40°41'50" j8 = 57°35'10" /. 436 402-7 = 1256-3 d- 0*263 88 d 2 496 SURVEYING PROBLEMS AND SOLUTIONS This is a quadratic equation in d 0-263 88 d z - 1256-3 d+ 436402-7 = d = 1256-3 ± N /(1256-3 2 - 4 x 0-26388 x 436 402-7) 2x0-26388 1256-3 ±^(1578 289-7 - 460631-8) 2x0-26388 1256-3 ±V 1117657-9 2x0-26388 1256-3 + 1057-2 2313-5 199-1 or 2x0-26388 2x0-26388 2x0-26388 The first answer is not in accordance with the data given. .-. d = 377-26 m /. Co-ordinates of E = 0, and -377-26 cosec 57°35'10" = Total Pep. Total Lat. -446-9 m Length EF = AG - HG + EJ = 1256*3- 377*26 cot a +377*26 cot j8 = 1256-3 + 377-26 (cot )8- cot a) = 1256-3 - 199-1 = 1057-2 Co-ordinates of F: P. Dep. 1057-2 sin 57°35'10" = +892-5 m P. Lat. 1057-2 cos 57°35'10" = +566*7 m Total Dep. of F = + 892*5 = +892-5 m Total Lat. of F = -446*9 + 566-7 = +119-8 m Example 9.18 Given the previous co-ordinate values let the dividing line pass through a point whose co-ordinates are (703*8, 0). From previous information, AD is N 56°47'40" E, 1276*9 m DC is S 16°53'20" W AD sin D 1276*9 sin 39°54'20" ocr _ In triangle ADG, AG = ^- = ; — - — = 855-9 m sinG sinl06°53'20" AREAS 497 39° 54' 20" <J>=106 , 53'20" Fig. 9.44 AH = 703-8 (due E) HG = 855-9 - 703-8 = 152-1 m Area = -AD. AG sin A = \ x 1276-9 x 855-9 sin 33°12'20" = 299 268 m 2 Now Area ADFE = ~ Area ABCD = AADF+ AAHE- AHFG : &AHE-AHFG = 447488-299 268 i.e. by Eq.9.57, 148 220 = AH' - 447 488 m 2 = 148 220 m' HG 2 2(cot a + cot 6 ) 2(cot a + cot <f> ) As 6 = 90°, 296440 cot a(cot a + cot <f>) = AH 2 (cota+ cot0) - HG 2 cota i.e. 296440cot 2 a+cota[296 440cot<£-AH 2 +HG 2 ]-4tf 2 cot<£ = thus 296440 cot 2 a- 562 200 cot a + 150 388 = Solving the quadratic gives a = 32°25' The co-ordinates of E are thus 0, and 703-8 tan32°25' i.e. (0, -446-9 m) Exercises 9 1. In the course of a chain survey, three survey lines forming the sides of a triangle were measured as follows: 498 SURVEYING PROBLEMS AND SOLUTIONS Line Length (links) Inclination AB 570 level BC 310 1 in 10 CA 495 7° On checking the chain after the survey, it was found that its length was 101 links. Calculate the correct plan area of the triangle. (E.M.E.U. Ans. 0-77249 acres) 2. A piece of ground has been surveyed with a Gunter's chain with the following results in chains: AB 11*50, CA 8-26, DB 10-30, CE 12-47, BC 12-20, CD 9-38, DE 6-63. Calculate the area in acres. Subsequently it was found that the chain was 0-01 chain too long. Find the discrepancy in the previous calculation and indicate its sign. (L.U. Ans. 12-320 acres; -0*248 acres) 3. Undernoted are data relating to three sides of an enclosure, AB, BC and CD respectively, and a line joining the points D and A. Line Azimuth Length (ft) AB 010°00' 541-6 BC 088°55' 346-9 CD 159° 19' 601*8 DA 272°01' 654*0 The fourth side of the enclosure is an arc of a circle to the south of DA, and the perpendicular distance from the point of bisection of the chord DA to the curve is 132*6 ft. Calculate the area of the en- closure in acres. (Ans. 7*66 acres) 4. Plot to a scale of 40 inches to 1 mile, a square representing 2H acres. By construction, draw an equilateral triangle of the same area and check the plotting by calculation. (Ans. Side of square 2*5 in. Side of triangle 3-8 in.) 5. The following offsets 15 ft apart were measured from a chain line to an irregular boundary: 23-8, 18-6, 14-2, 16-0, 21*4, 30*4, 29*6, 24*2 ft. Calculate the area in acres. (Ans. 0*0531 acres) 6. Find the area in square yards enclosed by the straight line bound- aries joining the points ABCDEFA whose co-ordinates are: AREAS 499 Eastings (ft) Northings (ft) A 250 75 B 550 175 C 700 425 D 675 675 E 450 675 F 150 425 (R.I.C.S. Ans. 24791-6 yd 2 ) 7. The following table gives the co-ordinates in feet of points on the perimeter of an enclosed area A BCDEF. Calculate the area of the land enclosed therein. Give your answer in statute acres and roods. Point Departure Latitude + - + A 74-7 105-2 B 63-7 261-4 C 305-0 74-5 D 132-4 140-4 E 54-5 192-4 F 571-9 108-3 (R.I.C.S. Ans. 4 acres 0-5 roods) 8. Using the data given in the traverse table below, compute the area in acres contained in the figure A BCD EA. Side Latitude (ft) Departure (ft) AB + 1327 -758 BC + 766 + 805 CD -952 +987 DE -1949 + 537 EA +808 -1572 (I.C.E. Ans. 73-3 acres) 9. State in square inches and decimals thereof what an area of 10 acres would be represented by, on each of three plans drawn to scale of (a) 1 inch = 2 chains (b) 1/2500 and (c) 6 in. = 1 mile. (Ans. (a) 25 in 2 ; (b) 10-04 in 2 ; (c) 0-562 in 2 ). 10. State what is meant by the term 'zero circle' when used in con- nection with the planimeter. A planimeter reading tens of square inches is handed to you to enable you to measure certain areas on plans drawn to scales of (a) 1/360 (b) 2 chains to 1 inch (c) 1/2500 (d) 6 in. to 1 mile and (e) 40 in. to 1 mile. State the multiplying factor you would use in each 500 SURVEYING PROBLEMS AND SOLUTIONS instance to convert the instrumental readings into acres. (Ans. (a) 0-2066 (b) 4-0 (c) 9-96 (d) 177-78 (e) 4-0) 11. The following data relate to a closed traverse: Line Azimuth Length (m) AB 241°30'00" 301-5 BC 149°27'00" 145-2 CD 034°20'30" 415-7 DE 079°18'00" 800-9 Calculate (a) the 1 ength and beari tig of the line £ est 30", (b) the area of the figure A BCDEA, (c) the length and bearing of the line BX which will divide the area into two equal parts, (d) the length of a line XY of bearing 068°50' which will divide the area into two equal parts. (Ans. (a) 307°54' ; (b) 89 290 m 2 ; (c) 526-4 m 091°13'48"; (d) 407-7 m) Bibliography HADDOCK, M.H., The Basis of Mine Surveying (Chapman & Hall). CLENDINNING, J., The Principles of Surveying (Blackie). McADAM, R., Colliery Surveying (Oliver and Boyd). MIDDLETON, R.E. and CHADWICK, o., A Treatise on Surveying (Spon). THOMAS, W.N., Surveying (Edward Arnold). SALMON, V.G., Practical Surveying and Fieldwork (Griffin). PARRY, R. and JENKINS, W.R., Land Surveying (Estates Gazette). DA VIES, R.E., FOOTE, F.s. and KELLY, J.W., Surveying: Theory and Practice (McGraw-Hill). MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.). SMIRNOFF, M.V., Measurements for Engineering and Other Surveys (Prentice-Hall). 10 VOLUMES 10.1 Volumes of Regular Solids The following is a summary of the most important formulae. Prism (Fig. 10.1) V = cross-sectional area x perpendicular height i.e. V = Ah = A y h, (10.1) Fig. 10.1 a b * 1 — V Hi (a) (b) Cylinder (Fig. 10.2) This is a special case of the prism. V = Ah = 7Tr 2 h (10.2) A = 1/4 rrab (a) Fig. 10.2 In Fig. 10.2 (b) the cylinder is cut obliquely and thus the end area becomes an ellipse, i.e. V = Ah. V = —rrabh 4 (10.3) = -^ Troth (as b = 2r) = v4,/i, = nr z h % (10.4) (10.5) 502 SURVEYING PROBLEMS AND SOLUTIONS Pyramid (Fig. 10.3) V = — base area x perpendicular height = 3 Ah (10.6) A — area of rectangle A « area of triangle Fig. 10.3 Cone (Fig. 10.4) This is a special case of the pyramid. V = \-Ah = \iTT z h 3 3 (10.7) A = JT/-2 (circle) A = 'A rrao( ellipse) Fig. 10.4 In Fig. 10.4(b) the base is in the form of an ellipse. V = — nabh = -T-rrarh (as fe = 2r) (10.8) (10.9) Frustum of Pyramid (Fig. 10.5) V = ^(A + B + y/M) where h = perpendicular height A and B = areas of larger and smaller ends respectively. (10.10) VOLUMES 503 Fig. 10.5 Frustum of Cone (Fig. 10.6) This is a special case of the frustum of the pyramid in which A = ttR 2 , B = TTr*. •'• V = ^bTR Z +Trr 2 +y/7TR Z 7Tr Z ] = y[K 2 +r 2 +/?r] (10.11) Wedge (Fig. 10.7) Fig. 10.7 f^ J 504 SURVEYING PROBLEMS AND SOLUTIONS V = Sum of parallel edges x width of base x -?■ perpendicular height. ° i.e. wh (x + y + z) (10.12) The above formulae relating to the pyramid are proved as follows (Fig. 10.8). Area 6 Area of midsection M Area A Fig. 10.8 Let A = the base area h A = the perpendicular height B = the area of any section parallel to the base and at a perpen- dicular distance h a from the vertex. Then 8V = B8h but A B = h A h% .-. B Ah% n A .'. V = A h A J h% dh B = A #A X hi 3 1 " 3 Ah A In the case of the frustum (h A~ h B = h), (Eq. 10.6) VOLUMES 505 V -£J* dhi A = TJT iK ~ h B^(- h A + h A h B + h B ) 3/z h 3 A + Ah, Ah\ (10.13) But hi B = Ahl and y/B y/A •'• V = -[A + y/AB + B] (Eq. 10.10) If C is the area of the mid-section of the pyramid, then £__V27 = ^_ A h A 4h% .'. A = 4C . .. Ah A (A + 4C)h A •■ V = -~ = — -^ (10.14) o O Similarly, if M is the area of the mid-section between A and B, then M _ ii(h A + h B )\ 2 _ (h A + h B f . 4M = A(h*+2h A h B + h%) h A Ha = A + 24^ + B -r- = 1(4M- 4 - B) 506 SURVEYING PROBLEMS AND SOLUTIONS Substituting this value in Eq. (10.13), V = -[A + 4M+B] (10.15) The Prismoidal Formula From Eq. 10.15 it is seen that the volumes of regular solids can be expressed by the same formula, viz. the volume is equal to the sum of the two parallel end areas + four times the area of the mid-section x 1/6 the perpendicular height, i.e. = ^ [A + AM + B] 6 (Eq. 10.15) This formula is normally associated with the prismoid which is defined as 'a solid having two parallel end areas A and B, which may be of any shape, provided that the surfaces joining their perimeters are capable of being generated by straight lines.' N.B. The mean area is derived from the average of the corresponding dimensions of the two end areas but not by taking the average of A and B. *?** * ■V Fig. 10.9 Area M Area B VOLUMES 507 Newton's proof of this formula is to take any point X on the mid- section and join it to all twelve vertices of the three sections, Fig. 10.9. The total volume then becomes the sum of the ten pyramids so formed. This formula is similarly applicable to the cone and sphere. The cone (Fig. 10.10) = 2irr 2 h = -7rr 2 h (Eq.10.7) Area 6 Area Area A*trr' "-■«/? Fig. 10.10 The sphere (Fig. 10.11) V = ^[0 + 477r 2 +0] 6 Area B m = r r (10.16) Area M + - rrr z Area A—0 Fig. 10.11 N.B. The relative volumes of a cone, sphere and cylinder, all of the same diameter and height, are respectively 1, 2 and 3, Fig. 10.12. 2r Cone = — x rrr 2 3 = -nr' 3 (10.17) Sphere = -ttt* (Eq. 10.16) Cylinder = 2r x nr z = 2rrrf (10.18) Fig. 10.12 508 SURVEYING PROBLEMS AND SOLUTIONS Applying the prismoidal formula to the frustum of a cone, -![-*w + *,(*-f] V 77" h = — [R 2 +r z +R 2 +2Rr+r 2 ] 6 rrh [R 2 +r 2 +Rr] (Eq. 10.11) Applying the prismoidal formula to the wedge, v -6l2 ( * +y) + 4 (-r + _ r)4 +0 J wh[ x y x z z y *J = TL2 + 2 + 2 + 2 + 2 + 2j — ix + y + z\ 6 (Eq. 10.12) It thus becomes very apparent that of all the mensuration formulae the PRISMOIDAL is the most important. If in any solid having an x axis the areas (A) normal to this axis can be expressed in the form A = bx 2 + ex + d then the prismoidal formula applies precisely. The sphere may be regarded as made up of an infinite number of small pyramids whose apexes meet at the centre of the sphere. The heights of these pyramids are then equal to the radius of the sphere. r Volume of each pyramid = area of base x - x Volume of sphere = surface area of sphere x — Surface area of sphere = Sector of a sphere (Fig. 10.13) This is a cone OAC with a sphe- rical cap. ABC. The volume can be derived from the above arguments: Volume of sector = (curved surface area of segment) x r/3 volume of sphere l/3r = 4nr 2 (10.19) VOLUMES 509 (10.20) (by Eq.9.33) = Irrrxhx^ 2 = —nr z h J Segment of sphere This is the sector less the cone OAC V = —7rr 2 h - r^Tw 2 (r - h) But w z = r 2 -(r-/i) 2 = r z -r z +2rh-h z 2 , 1 V = -7rr z h- -7T(2rh-h z )(r-h) 2 2 1 = ^-7Tr z h -—7Tr 2 h + irrh 2 — nh 3 3 3 3 = -7Th z (3r-h) (10.21) 10.2 Mineral Quantities Flat seams The general formula for calculating tonnage is: plan area (ft 2 ) x thickness (ft) x 62'5 x S.G. Tonnage - — tons ZZW (10.22) Here 62-5 ~ the weight of 1 ft 3 of water in pounds S.G. = the specific gravity of the mineral. This gives the tonnage in a seam before working and takes no account of losses. Taking coal as a typical mineral, alternative calculations may be made. Tonnage = plan area (acres) x thickness (in.) x 101 S.G. (10.23) Here 1 acre of water 1 inch thick weighs approximately 101 tons. When the specific gravity of coal is not known, either (a) Assume S.G. of 1-25 - 1*3. (b) Assume 125 tons per inch/ acre 1250 - 1500 tons per foot/ acre. (c) Assume 1 yd 3 of coal weighs 0*9 - 1*0 ton, or (d) Assume 1 ft 3 of coal weighs 80 lb. For loss of tonnage compared with the 'solid' estimate assume 15-20%. 510 SURVEYING PROBLEMS AND SOLUTIONS Based on the International System (S.I.) units, the following con- version factors are required: lton = 1016-05 kg 1 cwt = 50-802 3 kg 1 lb = 0-453 592 37 kg 1 ft = 0-3048 m 1 ft 2 = 0-092 903 m 2 1 acre = 4046-86 m 2 1 ft 3 = 0-028 316 m 3 lyd 3 = 0-764 555 m 3 The weight of water is 1 g/cm 3 at 4°C (i.e. 1000 kg/m 3 ) /. Coal weighs ^ 1250 - 1300 kg/m 3 ( x 1000 kg/yd 3 ) 1 gallon = 4-546 09 litres = 0-004 546m 3 Inclined seams The tonnage may be obtained by using either (a) the inclined area or (b) the vertical thickness, i.e. V = A.fseca (10.24) where V = plan area t = thickness a = angle of inclination of full dip of seam Example 10.1 In a pillar and stall, or stoop and room workings, the stalls or rooms are 12 ft in width and the pillars are 40 yd square. Calculate the approximate tonnage of coal extracted from the stalls or rooms, in a seam 7 ft 9 in. in thickness, dipping 19° from the horizon- tal, under a surface area V/2 acres in extent. Assume a yield of 125 tons per inch-acre, and deduct 3%% for loss in working. (M.Q.B./M ) 1 1 L Stall 120' Pillar 120' 1 Fig. 10.14 VOLUMES 511 In Fig. 10.14, Total Area = (120 + 12) 2 Pillar Area = 120 2 /120\ 2 .'. % extraction = 10 °-[7^j x 10 ° ■■»'■- -inr (-(i = 17-36% Plan area of extraction = 1*5 x 17*36/100 acres Inclined area of extraction = 1*5 x (17*36/100) x sec 19° Volume of coal extracted = 1*5 x (17*36/100) sec 19 x 7*75 x 12 x 125 = 3201*5 tons , , 3201*5 x 15 _ fti Loss of volume = --rrr = 120 tons Approximate tonnage extracted = 3200 - 120 = 3080 tons Exercises 10(a) (Regular solids) 1. A circular shaft is being lined with concrete, of average thickness 18 in. The finished inside diameter is 22 ft and a length of 60 ft is being walled. In addition 23 yd 3 of concrete will be required for a walling curb. If each yd 3 of finished concrete requires (a) 700 lb cement (b) 1600 lb sand and (c) 2 500 lb aggregate, find, to the nearest ton, the quantity of each material required to carry out the operation. (Ans. 84 tons cement; 192 tons sand; 300 tons aggregate) 2. A colliery reservoir, circular in shape, with sides sloping at a uniform gradient and lined with concrete is to be constructed on level ground to the undernoted inside dimensions: Top diameter 40 m Bottom diameter 36 m Depth 9 m The excavation is to be circular, 42 m in diameter, with vertical sides 10*5 m deep. Calculate the volume of concrete required. (Ans. 122*63 m 3 ) 3. Two shafts — one circular of 20 ft diameter, and the other rectan- gular 20 ft by 10 ft -are to be sunk to a depth of 710 yd. The mate- rial excavated is to be deposited in the form of a truncated cone with- in an area of level ground 100 yd square. If the top of the heap is to 512 SURVEYING PROBLEMS AND SOLUTIONS be level and the angle of repose of the material 35°, what will be the ultimate height of the heap with the diameter at its maximum ? Assume the proportion of broken to unbroken strata to be 5 to 3 by volume (take 77= (22/7)). (Ans. 38-9 ft) 4. An auxiliary water tank in the form of a cylinder with hemispheri- cal ends is placed with its long axis horizontal. The internal dimen- sions of the tank are (i) length of cylindrical portion 24 m (ii) dia- meter 5 m (iii) overall length 29 m. Calculate (a) the volume of the tank and (b) the amount of water it contains to the nearest 100 litres when filled to a depth of 1 *07 m . (Ans. (a) 536-69 m 3 ; (b) 11400 litres) 5. Two horizontal drifts of circular cross-section and 16 ft excava- ted diameter cross each other at right-angles and on the same level. Calculate the volume of excavation in ft 3 which is common to both drifts ' (M.Q.B./S Ans. 2731 ft 3 ) 6. The plan of a certain building on level ground is a square with sides 200 ft in length for which mineral support is about to be acqui- red. The south side of the building is parallel to the line of strike of the seam, the full dip of which is due South at the rate of 12 in. to the yard. The floor of the seam is 360 yd under the surface at the centre of the building. Draw a plan of the building and protecting block to a scale of 1 in = 200 ft, allowing a lateral margin equal to one third of the depth of the seam at the edge of the protecting block opposite the nearest point of the protected area. Thereafter calculate the tonnage of coal contain- ed in the protecting block, the seam thickness being 70 in. and the sp. gr. 1*26. (M.Q.B./S Ans. 182860 tons) 7. A solid pier is to have a level top surface 20 ft wide. The sides are to have a batter of 2 vertical to 1 horizontal and the seaward end is to be vertical and perpendicular to the pier axis. It is to be built on a rock stratum with a uniform slope of 1 in 24, the direction of this maximum slope making an angle whose tangent is 0*75 with the direc- tion of the pier. If the maximum height of the pier is to be 20 ft above the rock, diminishing to zero at the landward end, calculate the volume of material required. (L.U. Ans. 160000 ft 3 ) 8. A piece of ground has a uniform slope North and South of 1 verti- cal to 20 horizontal. A flat area 200 ft by 80 ft is to be made by cut- ting and filling, the two volumes being equal. Compare the volumes of VOLUMES 513 excavation if the 200 ft runs (a) North and South (b) East and West. The side slopes are to be 1 vertical to 2 horizontal. (L.U. Ans. (a) 24 300 ft 3 ; (b) 4453 ft 3 ) 10.3 Earthwork Calculations There are three general methods of calculating volumes, which use (1) cross-sectional areas, (2) contours, (3) spot heights. 10.31 Calculation of volumes from cross-sectional areas In this method cross-sections are taken at right-angles to some convenient base line which generally runs longitudinally through the earthworks. The method is specifically applicable to transport cons- tructions such as roads, railways and canals but may be applied to any irregular volume. The cross-sectional areas may be irregular and thus demand the use of one of the previously discussed methods, but in many transport constructions the areas conform to various typical shapes, viz. sections (a) without crossfall, (b) with crossfall, (c) with part cut and part fill, (d) with variable crossfall. (a) Sections without crossfall, i.e. surface level (Fig. 10.15) n Formation peg 3 (b) Embankment Fig. 10.15 Sections without crossfall The sections may be cuttings or embankments but in either case the following terms are used: Formation width (w) Formation height Qi ), measured on centre line (<L) 514 SURVEYING PROBLEMS AND SOLUTIONS Side width (W) , for the fixing of formation pegs, measured from centre line. Side slopes or batter 1 in m, i.e. 1 vertical to m horizontal w Thus W = — + mh Cross-sectional area = ^ (w + 2W) = -yCw + vv + 2mh ) A = h (w + mh Q ) (10.25) (10.26) Example 10.2 A cutting formed in level ground is to have a forma- tion width of 40 ft (12- 19 m) with the sides battering at 1 in 3. If the formation height is 10 ft (3*05 m) find (a) the side width, (b) the cross- sectional area. Fig. 10.16 3x10' 10' 20' S* Here W = 40 f t (12*19 m) h = 10 ft (3-05 m) m = 3 w W = - + mh = 20 + 3 x 10 = 50_ft (15*24 m) Area A = h (w + mh ) = 10(40 + 30) = 700 ft 2 (65-03 m 2 ) The metric values are shown in brackets. (b) Sections with crossfall of 1 in k (often referred to as a two- level section) In both the cutting and embankment the total area is made up of three parts, Fig. 10.17. (1) Triangle AHB Area = &M, (2) Trapezium BHFD Area = h w (3) Triangle DFE Area = l /2h 2 d 2 . VOLUMES 515 * -\ (a) Cutting (b) Embankment Fig. 10.17 Sections with cross fall In both figures, h. ~ h Q _ ^ 2k h, = h o + w 2fc By the rate of approach method(see p. 432) ft, h^mk <*i = 1 1 k + m m k (10.27) (10.28) (10.29) 516 SURVEYING PROBLEMS AND SOLUTIONS ««j j ^2 h 2 mk and d 2 = —±- = -i— (10.30) (10.31) "2 1 1 m k k - m ••■ H\ = w 2 + d ' 2 fc + m W 2 = 2 k - m Total area = ~Mi + K w + 2 hzdz (10.32) (10.33) The area of the cross-section is best solved by working from first principles, but if the work is extensive a complete formula may be re- quired. Given the initial information as w, h , m and k, substitution of these values into the various steps gives, from Eq. (10.33), ( h °~lk) mk ( h ° + 2r) mk A =j t*L — + ± f~ + wh 2(Jk + m) 2(fc - ro) - u — + wh n 2(k 2 - m 2 ) mk[2H 2 k + 2(-) + -£-J 2(*- - m 2 ) + Who . " [*»'*' + Q + ^ m \ ^ (10 .34) fc 2 - m z + W ° Example 10.3 The ground slopes at 1 in 20 at right-angles to the centre line of a proposed embankment which is to be 40 ft (12* 19 m) wide at a formation level of 10 ft (3*05 m) above the ground. If the batter of the sides is 1 in 2, calculate (a) the side width, (b) the area of the cross-section. In Fig. 10.18, w = 40 ft (12-19 m) h = 10 ft (3-05 m) VOLUMES 517 Then 771 = 2 k = 20 «*i d, = 9 x 2 x 20 360 = — = 16-36 ft 20 + 2 22 11 x 2 x 20 440 = 24-44 ft 20-2 18 W, = 20 + 16-36 = 36-36 ft (11-08 m) W 2 = 20 + 24-44 = 44-44 ft (13*55 m) Area = Vilh^ + fe 2 d^] + wh = l / 2 [9 x 16-36 + 11 x 24-44] + 40 x 10 = V 2 [147-28 + 268-88] + 400 = 608-08 ft 2 (56-49 m 2 ) By Eq. (10.34), 2[100 x 400 + 400 + 40 x 10 x 2] ^4 _ . . — + 40 x 10 400 - 4 40000 + 400 + 800 198 = 608-08 ft 2 + 400 or converted into S.I. units, 2[3-05 2 x 400 + 6-095 2 + 12-19 x 3-05 x 2] A = 396 [3721 + 37-15 + 74-36] + 37>18 + 12-19 x 3-05 198 = 19-36 + 37-18 56-54 m : 518 SURVEYING PROBLEMS AND SOLUTIONS (c) Sections with part cut and part fill (Fig. 10.19) Fig. 10. 19 Section part cut/part fill As before, the formation width BD = w the formation height FG = h the ground slope = 1 in k but here the batter on the cut and the fill may differ, so batter of fill is 1 in n batter of cut is 1 in m. The total area is made up of only 2 parts: (1) Triangle ABC Area = l / 2 h,d, (2) Triangle CED Area = y i h z d z w w <*, = 2 ~ X = 2 " **■ w w d 2 =- + x = —+kh (10.35) (10.36) By the rate of approach method and noting that /i, and h z are now required, it will be seen that to conform to the basic figure of the method the gradients must be transformed into n in 1, m in 1 and k in 1. and h } = K = <*, k - n k - m w Side width W. = - + HB ' 2 - | + nh, (10.37) (10.38) (10.39) VOLUMES 519 w W 2 = - + DJ w + mh 2 (10.40) 2 Area of fill = Vih,d % *,' 2(k - n) 2(fc - n) (10.41) Area of cut = yih 2 d z •k- %k - m) 2(fc - m) In the above ft has been treated as -ve, occurring in the cut. If it is +ve and the centre line is in fill, then (10.42) (? + **•)' Area of fill = K - (10.43) 2(/c - n) /w \ 2 Area of cut = (10.44) 2(fc - m) N.B. If h = and m = «, w 2 Area of cut = Area of fill = — (10.45) &(k - m) Example 10.4 A proposed road is to have a formation width of 40 feet with side slopes of 1 in 1 in cut and 1 in 2 in fill. The ground falls at 1 in 3 at right-angles to the centre line which has a reduced level of 260*3 ft. If the reduced level of the road is to be 262*8 ft, calcu- late (a) the side width, (b) the area of cut,(c) the area of fill. w w 4 = 2 - <-M) = 2 + ok = 20 + 2*5 x 3 = 27*5 ft 520 SURVEYING PROBLEMS AND SOLUTIONS W, = 75-0" W 2 = 26-5' « 5 *. + 262-8 +260-3 Fig. 10.20 d 2 = 20 - 7-5 = 12-5 ft h. *i 27-5 "3-2 = 27-5 ft k - n 1u d z 12-5 = 6-25 ft k — m w W, = - + nh, = 20 + 2 x 27*5 = 75-0 ft W 2 = - + mh 2 = 20 + 1 x 6-5 = 26'5 ft Area of cut = l / 2 d 2 h 2 = % x 12*5 x 6*25 = 39-06 ft 2 Area of fill = 1 / 2 d,ft, = Vi x 27*5 x 27-5 = 378*13 ft 2 By Eqs. 10.43/10.44, .2 Area of cut = Area of fill = G- tt °) 2(k-m) (=♦«.)' 2(k - n) (20 - 7-5) 2(3 - 1) (20 + 7-5)' 2(3 - 2) = 39-06 ft : = 378-13 ft 2 Example 10.5 An access road to a small mine is to be constructed to rise at 1 in 20 across a hillside having a maximum slope of 1 in 10. The road is to have a formation width of 15 ft, and the volumes of cut and fill are to be equalised. Find the width of cutting, and the volume of excavation in 100 ft of road. Side slopes are to batter at 1 in 1 in cut and 1 in 2 in fill. (N.R.C.T.) To find the transverse slope (see page 413) VOLUMES 521 Fig. 10.21 Let AB be the proposed road dipping at 1 in 20 (20 units) AC the full dip 1 in 10 (10 units) AD the transverse slope 1 in t (t units). In triangle ABC, COS0 = 10 In triangle ADC, = 11*55 (gradient value) Fig. 10.22 If area of cut = area of fill, from Eqs. (10.43) and (10.44) for h +ve, (}-k„) 2 d + 4 2(k - m) 2{k - n) (7-5 - ll'SShf (7-5 + ll-SShf i.e. 11-55 - 1 7-5 - ll-55/i V 9. 11-55 - 2 55 (7-5 + ll-55/i) 9-55 = 1-051(7-5 + ll-55fc) — 0*383 H = 13^89 = - '° 1617 <*■«•*"*> x = kh = 11-55 x 0-01617 = -0-187 ft Width of cutting = 7-5 + 0-187 = 7-687 say 7*69 ft 522 SURVEYING PROBLEMS AND SOLUTIONS Area of cutting (7-5 + 0-187) ! 2(11-55 - 1) 7-69 2 21-10 2-80 ft 2 Volume of cutting = 2-80 x 100 ft 3 = 10-37 yd 3 (d) Sections with variable crossfall (three-level section) If the cross-section is very variable, it may be necessary to deter- mine the area either (a) by an ordinate method or (b) by plotting the section and obtaining the area by scaling or by planimeter. If the section changes ground slope at the centre line the follow- ing analysis can be applied, Fig. 10.23. Fig. 10.23 Section with variable crossfall The total area is made up of four parts: (1) Triangle AHB Area = i-h } d, (2) Trapezium BHGC Area = ^{h + fc, ) (3) Trapezium CGFD Area = j(h + h 2 ) (4) Triangle DFE Area = !fc 2 d 2 Total Area = \[h,d,+ h 2 a\ + |(2fc + /i, + h 2 )] w Here fc, = h - - K = K + w 27 <*, = h x mk k + m (10.46) VOLUMES 523 , _ hzinl 2 = 7^- Side width *» w + 4 w ». - 2 + 4 N.B. fc and / have both been assumed +ve, and the appropriate change in sign will be required if GA and GE are different from that shown. // the level of the surface is known, relative to the formation level, at the edges of the cutting or embankment (Fig. 10.24) Fig. 10.24 Section with levels at formation pegs Area ABCDHA = Area XBDH - Area XBA = J (f + <*,)(#, + h ) -H,d] Area DEFGH = Area DFYH - Area FYG Total Area - m(fi* + H 2 2 )j. ,) (10.47) Exercises 10(b) (Cross-sectional areas) 9. At a point A on the surface of ground dipping uniformly due South 1 in 3, excavation is about to commence to form a short cutting for a branch railway bearing N 30° E and rising at 1 in 60 from A . The 524 SURVEYING PROBLEMS AND SOLUTIONS width at formation level is 20 ft and the sides batter at 1 vertical to 1 horizontal. Plot two cross-sections at points B and C 100 ft and 150 ft respec- tively from A and calculate the cross-sectional area at B. (N.R.C.T. Ans. 1323* 3 ft 2 ) 10. Calculate the side widths and cross-sectional area of an em- bankment to a road with a formation width of 40 ft. The sides slope 1 in 2 when the centre height is 10 ft and the existing ground has a crossfall of 1 in 12 at right-angles to the centre line of the embank- ment. (N.R.C.T. Ans. 34*28 ft; 48'01 ft; 622*8 ft 2 ) 11. A road is to be constructed on the side of a hill having a cross- fall of 1 vertically to 8 horizontally at right-angles to the centre line of the road; the side slopes are to be similarly 1 to 2 in cut and 1 to 3 in fill; the formation is 50 ft wide and level. Find the distance of the centre line of the road from the point of intersection of the forma- tion with the natural ground to give equality of cut and fill, ignoring any consideration of 'bulking'. (L.U. Ans. 1*14 ft on the fill side) 12. A road is to be constructed on the side of a hill having a cross- fall of 1 vertically to 10 horizontally at right-angles to the centre line of the road; the side slopes are to be similarly 1 to 2 in cut and 1 to 3 in fill; the formation is 80ft wide and level. Find the position of the centre line of the road with respect to the point of intersection of the formation and the natural ground, (a) to give equality of cut and fill, (b) so that the area of cut shall be 0*8 of the area of fill in order to allow for bulking. (L.U. Ans. (a) 1-34 ft on the fill side; (b) 0*90 ft on the cut side) 13. The earth embankment for a new road is to have a top width of 40 ft and side slopes of 1 vertically to 2 horizontally, the reduced level of the top surface being 100*0 O.D. At a certain cross-section, the chainages and reduced levels of the natural ground are as follows, the chainage of the centre line being zero, those on the left and right being treated as negative and positive respectively: Chainage (ft) -50 -30 -15 -0 +10 +44 Reduced level (ft) 86*6 88*6 89*2 90*0 90*7 92*4 Find the area of the cross- section of the filling to the nearest square foot, by calculation. (L.U. Ans. 567 ft 2 ) 14. A 100 ft length of earthwork volume for a proposed road has a constant cross-section of cut and fill, in which the cut area equals the VOLUMES 525 fill area. The level formation is 30 ft wide, the transverse ground slope is 20° and the side slopes in cut and fill are respectively ^(horizontal) to 1 (vertical) and 1 (horizontal) to 1 (vertical). Calculate the volume of excavation in 100 ft length. (L.U. Ans. 209-2 yd 3 ) 10.32 Alternative formulae for the calculation of volumes from the derived cross-sectional areas Having computed the areas of the cross-sections, the volumes in- volved in the construction can be computed by using one of the ordinate formulae but substituting the area of the cross-section for the ordinate. (1) Mean Area Rule W v = -04, +A 2+ A a + ... +AJ i.e. n (10.48) where W = total length between end sections measured along centre line. n = no. of sectional areas. 2-4 = sum of the sectional areas. N.B. This is not a very accurate method. (2) Trapezoidal (or End Area) Rule (Fig. 10.25) < A 2 "3 *3 >.'• r^ A - * . - *2 . . W* - * w * » - "* » . ** . . w ~ iw Fig. 10.25 Trapezoidal rule y z = T 2 (a z + a 3 ) w V 3 = ^(A 3+ A 4 ) 2 ■&4n-l + K) 526 SURVEYING PROBLEMS AND SOLUTIONS If W, = W 2 = W n , then V = (V, + V 2 + V 3 + ... +V n _,) w 2 = - U, + 2A 2 + 2A 3 + 2A A + ... +2A n _, + 4J (10.49) (3) Prismoidal Rule (Fig. 10.26) As the cross-sections are all parallel and the distance apart can be made equal, the alternate sections can be considered as the mid- section. The formula assumes that the mid-section is derived from the mean of all the linear dimensions of the end areas. This is difficult to apply in practice but the above application is considered justified particularly if the distance apart of the sections is kept small. '1 "3 *3 ^4 *4 M W - . w » Fig. 10.26 Prismoidal rale Thus (V, + V 2 ) = ^04, + 4A 2 + A 3 ) (V, + V 4 ) = |(^ 3 + 4A 4 + A s ) (V 8 + V 6 ) = |o4 5 + 4A 6 + A 7 ) w .% Total volume = -U, + 4A 2 + 2A 3 + 4A A + 2A S + 4A 6 + A 7 ] If the number of sections is odd, then w V = -3 [4, + 41 even areas + 22 odd areas + 4J (10.50) which is Simpson's rule applied to volumes. Prismoidal Corrections If having applied the end areas rule it is then required to find a closer approximation, a correction can be applied to change the deri- ved value into the amount that would have been derived had the pris- moidal rule been applied. For areas /I, and A 2 s units apart, By the end areas formula, V^, = —(,4, + A 2 ) VOLUMES 527 s, By the prismoidal formula, V P = t(^i + 4A m + A 2 ) The difference will be the value of the correction, i.e. = v e - V P = |[3A, + 3A 2 - A, - AA m - A 2 ] c 6 s c = H204, + A 2 ) - 4A m ] (10.51) 6 For sections without crossfall Let the two end sections A y and A 2 be s it apart with formation width of w ft and formation heights h r and h 2 . Then, by Eq. (10.26), A^ = A, (w + mh y ) A 2 = h 2 (w + mh 2 ) Putting these values into Eq. (10.51), c = I^IwCft, + h 2 ) + m(h* + h 2 )\ - 2\wQi, + h 2 ) + f (/if + hi + 2h,h 2 )\] = ™[2h? + 2hl - h z - hi - 2h<h 2 ] c = ~?U, - fc 2 ) 2 (10.52) 6 For sections with crossfall From Eq. (10.34), m\h z k z +— w z + wh.ml A % = — LJ _±__ !_ J + w /i, (fc 2 -m 2 ) m\h z k z + -w 2 + w/i,/m| 4, = — ^ 1 3 — J + wfc 2 2 (k 2 - m 2 ) m[i(/i 1 + fc 2 ) 2 k 2 + ^w 2 + >(ft, + h 2 )] , ^4 m = — — r- 5 — 5 +-w(h.+h 9 ) (fc 2 -ro 2 ) 2 ' 2 Substituting these values in Eq. (10.51), Prismoidal Correction c = ^[2C4, + A 2 ) - 4A m ] 528 SURVEYING PROBLEMS AND SOLUTIONS c = ar^ zP * fc W + n t ) + I w 2 + wm(h, + h z ) oC/c 4 - m ) + wm(k 2 - m z )(h x + h 2 ) \ -4{±k 2 (ht + h 2 ) 2 + {w* + ±wm(k 2 -m 2 )(h x +h 2 )}] c = 6( fe/ m m2) [2Jc 2 (ftf + fc 2 2 ) - /c 2 (/i, + fc 2 ) 2 ] sm/c (ft, - /i 2 ) 2 C 6(fc 2 - m 2 ) For sections with cut and fill From Eq. (10.42), (^ + kh\ (10.53) A. K -rim — 2(fc - m) (I + kh *) Z 2(k - m) 2(k - m) Substituting these values in Eq. (10.51), Prismoidal correction = f \ 2 l(™ + kh \ 2 + (l +kh \ 2 ) for cut 12(k-m)L \\2 7 V 2 7 J -«{H». ♦*.>)'] Prismoidal correction _ s [? / / * _ kh¥ + (- - kh \ 2 1 for fill = 12(fc -n)[ l\2 / \2 7 / = s/c 2 ^ - ft 2 ) 2 Q 12 (k - n) Example 10.6 An embankment is to be formed with its centre line on the surface (in the form of a plane) on full dip of 1 in 20. If the formation width is 40 ft and formation height are 10, 15, and 20 ft at intervals of 100 feet, with the side slopes 1 in 2, calculate the volume between the end sections. VOLUMES 529 Fig. 10.27 Longitudinal section 40' ^ Ay ho' |£1< j^ A, 20' Fig. 10.28 Cross-sections Area (1) = h y (yv + mh } ) = 10(40 + 2 x 10) = 600 ft 2 Area (2) - 15(40 + 2 x 15) = 1050 ft 2 Area (3) = 20(40 + 2 x 20) = 1600 ft 2 Volume (1) By Mean Areas W V = £<2i4) 200 (600 + 1050 + 1600) = 216 666-7 ft c 530 SURVEYING PROBLEMS AND SOLUTIONS (2) By End Areas (Trapezoidal) V = ^[A t + 2A 2 + A 3 ] = ^[600 + 2100 + 1600] = 215000 ft 3 (3) By the prismoidal rule (treating the whole as one prismoid) V = |U, + 4A 2 + A 3 ] = ™ [600 + 4200 + 1600] = 213 333-3 ft 3 (4) By Prismoidal Correction to End Areas (By end Areas _ 100 + _ 82m ft3 in each section) I V 2 = ^(1050 + 1500) = 132500 ft 3 = 215000 ft 2 By outer Areas V = -^(600 + 1600) = 220000 ft 3 Applying Prismoidal Correction to adjacent areas, <y E - v p\ = ^pOh -fc.) a = 10 ° x 2 (i 5 __ 10) 2 = 833-33 ft 3 o (V E - V P ) 2 = 10 ° 6 X 2 (20 - 15) 2 = 833-33 ft 3 Total Correction = + 1666-66 ft 3 .". Volume V p =215000 - 1666-67 = 213 333-33 ft 3 Applying Prismoidal Correction to outer areas, V e ~ V p = 2 °° 6 X 2 (20 - 10) 2 = 6666-67 V P = 220000-6666-67 = 213 333-33 ft 5 VOLUMES 531 N.B. (1) The correct value is only obtained by applying the prismoidal correction to volumes obtained by adjacent areas unless, as here, the whole figure is symmetrical. (2) The prismoidal correction has little to commend it in prefer- ence to the application of the prismoidal formula if all the information is readily available. Example 10.7 Given the previous example but with the centre line turned through 90° 40' Fig. 10.29 Cross-sections From Eq. (10.34), m[h 2 k z + -w 2 + whm] A = + wh kr - m z Cross-sectional Areas 2 , - 20^72^ [1 ° 2 x 20 2 + J- x 40 2 + 40 x 10 x 2] + 40 x 10 A. = 198 J. 198' 1 [40000 + 400 + 800] + 400 = 608*08 ft 2 K = ^[90000 + 400 + 1200] + 600 = 1062-62 ft 5 A„ = 198 [160000 + 400 + 1600] + 800 = 1618- 18 ft 2 Volume (1) By Mean Areas (Eq. 10.48) V = ^[608-08 + 1062-62 + 1618-18] = 219258-7 ft 3 (2) By End Areas (Eq. 10.49) (taking all y = 100 [60g . 08 106 2-62 + 1618-18] - 217575 ft 3 sections) 2 (taking outer y _ 2W [60g . 08 + „„.„, _ 222626 ft , sections) 2 532 SURVEYING PROBLEMS AND SOLUTIONS (3) By the prismoidal rule (Eq. 10.50) (treating the whole as one prismoid) V = ^ [608-08 + 4 x 1062-62 + 1618-18] = 215891-5 ft 3 (4) By applying Prismoidal Correction to End Areas Applying prismoidal correction to each section (Eq. 10.53), Smk 2 c = c 2 = 6(/r - to 2 ) 100 x 2 x 20 2 6 x 396 100 x 2 x 20 2 396 2^ x (ft, - h 2 ) x (15 - 10) 2 = 841-75 ft* x (20 - 15) 2 = 841-75 iV = 1683-50 .-. V P = 217 575 - 1683-5 = 215 891-5 ft 5 Applying prismoidal correction to outer areas, c . **.* 2 .?. 202 x (20 - 10) 2 . 6734 ft- 6 x 396 V P = 222626 - 6734 = 215892 ft 3 N.B. Where the figure is symmetrical the prismoidal correction again gives the same value, but it would be unwise to apply the latter method where the prismoids are long and the cross-sectional areas very vari- able, unless it is applied to each section in turn, as shown above. Example 10.8 A road has a formation width of 40 ft, and the side slopes are 1 in 1 in cut and 1 in 2 in fill. The ground slopes at 1 in 3 at right-angles to the centre line. Sections at 100 ft centres are found to have formation heights of + 1 ft, 0, and -2 ft respectively. Calculate the volumes of cut and fill over this length. Fig. 10.30 VOLUMES 533 Areas of cut From Eq. (10.44), (! - kh Y (h + ve) Area of Cut = x - 7f71 ^ 2(/c - m) (f - 3_x 1) (20 - 0) 2 A < ' 2(3 - 1) = 72 ' 25 ft2 A 2 = ^ — '— = 100 ft 2 4 .2 (ft _ ve) . - (2 ° t 3 X 2y - 169 ft' Areas o/ /i/Z From Eq. (10.43), (h + ve) Area of fill ••• K = (20 + 3) 2(3 - 2) A' z = (20 + 0) 2 2 (/z-ve) K = (20 - 6) 2 2 Volume of Cut (1) By Mean i4 reas 2(fc --n) = 264-5 ft 2 = 200 ft 2 98 ft : V = 2 25[ 72 -25 + 100 + 169] = 22750 ft 3 (2) By End Areas (taking all sections) V = l°-P-[72-25 + 2 x 100 + 169] = 22062-5 ft 3 (taking outer sections) V = ^[72-25 + 169] = 24 125 ft 3 (3) By the prismoidal rule (treating the whole as a prismoid) V = ^[72-25 + 4 x 100 + 169] = 21375 ft 3 534 SURVEYING PROBLEMS AND SOLUTIONS (4) By applying Prismoidal Correction to End Areas Applying prismoidal correction to each section of cut, c = Sk\h, - h 2 ) 2 12(k - m) ... c = 100 x 3 2 (1 - 0) 2 _ 37 . 5 , 1 12(3 - 1) ~ 6/ * " m 900 x (2 - Of m ft3 2 24 c T = 187-5 ft 3 .-. V P = 22062-5- 187-5 = 21875-0 ft 3 Applying prismoidal correction to outer areas, 900 (1 + 2) 2 „ n r _ c = L '— = 337.5 ft 3 24 .-. V P = 24125 - 337*5 = 23787-5 ft 3 Volumes of Fill (1) By Mean Areas V = -^[264-5 + 200 + 98] = 37500 ft 3 (2) By End Areas (taking all sections) V = ^[264-5 + 2 x 200 + 98] = 38 125 ft 3 (taking outer sections) V = ^[264-5 + 98] = 36250 ft 3 (3) By the prismoidal rule (treating the whole as a prismoid) V = y^264-5 + 4 x 200 + 98] = 38750 ft 3 (4) By applying Prismoidal Correction to End Areas Applying prismoidal correction to each section of fill, 100 x 3 2 x l 2 C ' 12(3 - 2) = 75 ft 3 900 x 2 2 C > = 12 = 300 ft 3 c T = 375 ft 3 V P = 38 125 - 375 = 37750 fta VOLUMES 535 Applying prismoidal correction to outer areas, c = Vp = 36250 - 675 = 35 575 ft 3 N.B. The prismoidal correction applied to each section gives a much closer approximation to the value derived by the prismoidal formula, although the latter in this case is not strictly correct as the middle height is not the mean of the two end heights. Example 10.9 Calculate the volume between three sections of a rail- way cutting. The formation width is 20 ft; the sections are 100 ft apart; the side slopes are 1 in 2 and the heights of the surface above the formation level are as follows: Section Left Centre Right 1 17-6 16-4 17-0 2 21-2 20-0 18-8 3 19-3 17-9 16-3 From Eq. (10.47), Area = 1 |f + nitty (H, + h ) + (| + mfty(H 2 + h ) - m(H? + H 2 * )j Section 1, A, = ± U y + 2 x 17-e) (17-6 + 16*4) +fy + 2x 17-o)(17-0 + 16-4) - 2(17-6 2 +17-0 2 ) = 904-4 ft 2 Section 2, A 2 = I [(10+ 42-4) (41*2) + (10 + 37-6) (38-8) - 2(21-2 2 + 18-8 2 )] = 1200-0 ft 2 Section 3, A 3 = j[(10 + 38-6)(37-2) + (10 + 32-6)(34-2) + 2(19-3 2 +16-3 2 )] = 994-2 ft 2 Using the prismoidal formula, V = 3^7t904-4 + 4x 1200 + 994-2] = 8270 yd 3 Total Volume = 8 270 yd 3 10.33 Curvature Correction (Fig. 10.31) When the centre line of the construction is curved, the cross- sectional areas will be no longer parallel but radial to the curve. 536 SURVEYING PROBLEMS AND SOLUTIONS Volume of such form is obtained by using the Theorem of Pappus which states that 'a volume swept out by a constant area revolving about a fixed axis is given by the product of the area and the length of the path of the centroid of the area*. The volume of earthworks involved in cuttings and embankments as part of transport systems following circular curves may thus be determined by considering cross-sectional areas revolving about the centre of such circular curves. Area A, Centroid Section at X Centre line of formation Centre of curvature & Centroid Section at Y Fig. 10.31 Curvature correction If the cross-sectional area is constant, then the volume will equal the product of this area and the length of the arc traced by the centroid. If the sections are not uniform, an approximate volume can be de- rived by considering a mean eccentric distance (e) = e * + ez relative to the centre line of the formation. This will give a mean radius for the path of the centroid (/? ± e), the negative sign being taken as on the same side as the centre of curvature. Length of path of centroid XY = (R + e) rad . but &ra.d = "o where S = length of arc on the centre line .-. XY = |(K + e) = s(l ± |) VOLUMES 537 Volume is given approximately as v = |(A + 4,)(i±f) (10.56) Alternatively each area may be corrected for the eccentricity of its centroid. If e, be the eccentricity of the centroid of an area A, , then the volume swept out through a small arc 8d is SV = A^(R ± e,)50. If the eccentricity had been neglected then 8V = A.R80 with a resulting error = A^e^bd _ Vi R per unit length (10.57) Ae Thus, if each area is corrected by an amount ±~H"> these new equivalent areas can be used in the volume formula adopted. 10.34 Derivation of the eccentricity e of the centroid G Centroids of simple shapes Parallelogram (Fig. 10.32) G lies on the intersection of the diagonals or the intersection of lines joining the midpoints of their opposite sides. (10.58) Fig. 10.32 Triangle (Fig. 10.33) G lies at t from each apex G lies at the intersection of the medians and is -=- of their length (10.59) 538 SURVEYING PROBLEMS AND SOLUTIONS Trapezium (Fig. 10.34) (10.60) Fig. 10.35 Fig. 10.34 A Compound Body (Fig. 10.35) If the areas of the separate parts are A x and A z and their cen troids G, and G z , with the compounded centroid G, G = A 2 xG,G 2 1 A, + A z A, x G,G 2 or G 2 G = ^i + A z (10.61) (10.62) Thus for typical cross-sectional areas met with in earthwork cal- culations, the figures can be divided into triangles and the centre of gravity derived from the compounding of the separate centroids of the triangles or trapezium, Fig. 10.36. Fig. 10.36 Alternatively, Fig. 10.37, Let the diagonals of ABCD intersect at E. BO = OD on line BD AE = FC on line AC then 2.0G = GF Fig. 10.37 (10.63) VOLUMES 539 To find the eccentricity e of the centroid G Case 1. Where the surface has no crossfall, the area is symmetrical and the centroid lies on the centre line, i.e. e - 0. Case 2. Where the surface has a crossfall 1 in k (Fig. 10.38) Let Total Area of ABDE = A T Area of triangle AEF *-«, = \AF(H 2 -H,) _*W+ W 2 ) (10.64) B I . wlz 7 F ° Fig. 10.38 Section with crossfall Let G, and G 2 be the centroids of areas AEF and AFDB respectively. Length AQ = horizontal projection of AG y m I pi? + AFl = i^ + ^ +2R;] = iUW, + WJ = W, + -^ Distance of Q from centre line, W, i.e. XQ, = G 2 g, - HJ +J18-- Hf, (10.65) Distance of centroid G for the whole figure (from the centre line, i.e. e), Area A 4EF x XQ ¥f y W Z (W, + W 2 ) e = Total Area A, 3k. A, (10.66) Conversion Area A c - ± A r e R 540 SURVEYING PROBLEMS AND SOLUTIONS i.e. A = + A T [H\W 2 (W, + W 2 )] W,W 2 (W, + W 2 ) Corrected Area = A T ± 3k R W,W 2 (W, + W 2 ) ZkR Case 3. Sections with part cut and part fill (Fig. 10.39) (10.67) < ^7 J e2 > f '« ^ ^ 2 Fig. 10.39 Section part cut/part fill For section in cut, i.e. triangle CED, G lies on the median EQ. J Q = I (f + *) ~ x = I ("f - *) e 2 = JQ + i(W 2 - JQ) = I(^ + 2JQ) -*(« w W 2 + j- *ft . ) Similarly for fill e, = j (W, + ^ + &O (10.68) (10.69) Example 10.10 Using the information in Example 10.6, viz. embank- ment with a surface crossfall of 1 in 20, side slopes 1 in 2, formation width 40 ft and formation heights of 10, 15 and 20 ft at 100 ft cent- res, if this formation lies with its centre line on the arc of a circle of radius 500 ft, calculate (a) the side widths of each section, (b) the eccentricity of their centroids, (c) the volume of the embankment over this length for the centre of curvature (i) uphill (ii) downhill. VOLUMES 541 Fig. 10.40 (a) Side widths Section 1 From Eq. (10.31), W, = (*• " I) mk w 2 k + m and from Eq. (10.32), W„ = (*° - fk) mk w 2 k - m 40 i.e. W f = -V + *» " D 2 x 20 2 ' 20+2 9 x 40 Section 2 Section 3 W, = w. = w„ = w. = w„ = 20 + 20 + 20 + 20 + 20 + 20 + 22 11 x 40 18 14 x 40 22 16 x 40 18 19 x 40 22 21 x 40 18 = 36-36 ft 44-44 ft 45-45 ft 55-56 ft 54-55 ft _ 66-67 ft (b) Eccentricity (e) From Eq. (10.66), e 3k A 542 SURVEYING PROBLEMS AND SOLUTIONS _ 36-36 x 44-44 (36-36 + 44-44) ?1 3 x 20 x 608-08 = 3-58 ft (Area 608*08 ft 2 from previous calcula- tions) = 45-45 x 55-56 (45-45 + 55-56) &z 3 x 20 x 1062-62 = 4-00 ft = 54-55 x 66-67 (54-55 x 66-67) 63 3 x 20 x 1618-18 = 4-54 ft (c) Volumes Using the above values of eccentricity in the prismoidal formula, the volume correction V . ± M[ 608 -08 x |f ♦ 4 (1062-62 , ^) + 1618-18 x±|] = ± ^[608-08 x 3-58 + 16 x 1062-62 + 1618-18 x 4-54] = ±1768-4 ft 3 The correction is + ve if the centre of the curve lies on the uphill side. .*. Corrected volume = 215891-5 ± 1768*4 ft 3 = 217 660 ft 3 or 214 123 ft 3 A more convenient calculation of volume, without separately cal- culating the eccentricity, is to correct the areas using Eq. (10.67). Area Correction A„ = ± • 2V ! 11. Z.k.R A - + 36 ' 36 x 44 ' 44 (36 ' 36 + 44 ' 44) °' " " 3 x 20 x 500 = ± 4-35 ft A = ± 45 ' 45 x 55 ' 56 ( 45 ' 45 + 55 ' 56 ) ° 2 3 x 20 x 500 = ± 8-50 ft 2 A = + 54-55 x 66-67 (54-55 + 66-67) ° 3 3 x 20 x 500 = ± 14-70 ft 2 VOLUMES 543 Corrected Areas are: A, = 608-08 ±4-35 = 612-43 or 603-73 ft 2 A 2 = 1062-62 ± 8-50 = 1071-12 or 1054-12 ft 2 A 3 = 1618-18 ±14-70 = 1632-88 or 1603-48 ft 2 Corrected volumes: (i) With centre of curve on uphill side, V - ^[612-43 + 4 x 1071-12 + 1632-88] - 217657 ft 3 (ii) With centre of curve on downhill side, V = ^[603-73 + 4 x 1054-12 + 1603-48] = 214122 ft 3 10.4 Calculation of Volumes from Contour Maps Here the volume is derived from the areas contained in the plane of the contour. For accurate determinations the contour interval must be kept to a minimum and this value will be the width (w) in the form- ulae previously discussed. The areas will generally be obtained by means of a planimeter, the latter tracing out the enclosing line of the contour. For most practical purposes the Prismoidal formula is satisfactory, with alternate areas as 'mid-areas' or, if the contour interval is large, on interpolated mid-contour giving the required 'mid-area' may be used. 10.5 Calculation of Volumes from Spot-heights This method uses grid levels from which the depth of construction is derived. The volume is computed from the mean depth of construction in each section forming a truncated prism, the end area of which may be rectangular but preferably triangular, Fig. 10.41. V = plan area x mean height (10.70) If a grid is used, the triangular prisms are formed by drawing diagonals, and then each prism is considered in turn. 544 SURVEYING PROBLEMS AND SOLUTIONS />, t 2 h \ / / \ / / s t / "4 \ ,'th • V / / \ N /»• "7 Fig. 10.41 Volume from spot-heights Fig. 10.42 The total volume is then derived (each triangle is of the same area) as one third of the area of the triangle multiplied by the sum of each height in turn multiplied by the number of applications of that height, i.e. V = ^[Inh] (10.71) e.g. V = -[2/i, + 2hz + 2h 3 + 2h 4 + 8h 5 + 2h 6 + 2hy + 2h e + 2h 9 ] (Fig. 10.42) 10.6 Mass-haul Diagrams These are used in planning the haulage of large volumes of earth- work for construction works in railway and trunk road projects. 10.61 Definitions Bulking An increase in volume of earthwork after excavation. Shrinkage A decrease in volume of earthwork after deposition and compaction. Haul Distance (d) The distance from the working face of the excava- tion to the tipping point. Average Haul Distance (D) The distance from the centre of gravity of the cutting to that of the filling. Free Haul Distance The distance, given in the Bill of Quantities, included in the price of excavation per cubic yard. Overhaul Distance The extra distance of transport of earthwork volumes beyond the Free Haul Distance. Haul The sum of the product of each load by its haul distance. This must equal the total volume of excavation multiplied by the average haul distance, i.e. S.v.d = V.D. VOLUMES 545 Overhaul The products of volumes by their respective overhaul dis- tance. Excess payment will depend upon overhaul. Station Yard A unit of overhaul, viz. 1 yd 3 x 100 ft. Borrow The volume of material brought into a section due to a deficiency. Waste The volume of material taken from a section due to excess. In S.I. units the haul will be in m 3 , the haul distances in metres and the new 'station' unit probably 1 m 3 moved 100 m. 10.62 Construction of the mass-haul diagram (Fig. 10.43) (1) Calculate the cross- sectional areas at given intervals along the project. (2) Calculate the volumes of cut and fill between the given areas relative to the proposed formation. N.B. (a) Volumes of cut are considered positive, (b) Volumes of fill are considered negative. (3) Calculate the aggregated algebraic volume for each section. (4) Plot the profile of the existing ground and the formation. (5) Using the same scale for the horizontal base line, plot the mass haul curve with the aggregated volumes as ordinates. M P Max. haulage point Mass haul diagram Fig. 10.43 Mass-haul curves 546 SURVEYING PROBLEMS AND SOLUTIONS 10.63 Characteristics of the mass-haul diagram (1) A rising curve indicates cutting as the aggregate volume is in- creasing (a—/ is seen to agree with AF on the profile). (2) A maximum point on the curve agrees with the end of the cut, i.e. f-F. (3) A falling curve indicates filling as the aggregate volume is decreasing (f-k is seen to agree with F-K on the profile). (4) The vertical difference between a maximum point and the next minimum point represents the volume of the embankment, i.e. //, + k,k (the vertical difference between any two points not having a minimum or maximum between them represents the volume of earthwork between them.) (5) If any horizontal line is drawn cutting the mass-haul curve (e.g. aqp), the volume of cut equals the volume of fill between these points. In each case the algebraic sum of the quantities must equal zero. (6) When the horizontal balancing line cuts the curve, the area above the line indicates that the earthwork volume must be moved for- ward. When the area cut off lies below the balancing line, then the earthwork must be moved backwards. (7) The length of the balancing line between intersection points, e.g. aq, qp, represents the maximum haul distance in that section (q is the maximum haulage point both forward, aq, and backwards, pq). (8) The area cut off by the balancing line represents the haul in that section. N.B. As the vertical and horizontal scales are different, i.e. 1 in. = s ft horizontally and 1 in. = v yd 3 , an area of a in 2 repre- sents a haul of avs yd 3 , ft = ^r-r station yards. 10.64 Free-haul and overhaul (Fig. 10.44) The Mass-haul diagram is used for finding the overhaul charge as follows: Free-haul distance is marked off parallel to the balance line on any haul area, e.g. bd. The ordinate cc 2 represents the volume dealt with as illustrated in the profile. Any cut within the section ABB^ 4, has to be transported through the free- haul length to be deposited in the section D^E^ED. This rep- resents the 'overhaul' of volume (ordinate bb,) which is moved from the centroid G, of the cut to the centroid G 2 of the fill. The overhaul distance is given as the distance between the cen- troids less the free-haul distance. i.e. (G t G 2 ) - bd VOLUMES 547 Balance line (b) Profile Fig. 10.44 Free-haul and overhaul The amount of overhaul is given as the volume (ordinate bb % = dd,) x the overhaul distance. Where long haulage distances are involved, it may be more eco- nomical to waste material from the excavation and to borrow from a location within the free-haul limit. If / is the overhaul distance, c the cost of overhaul and e the cost of excavation, then to move 1 yd 3 from cut to fill the cost is given as e + Ic whereas the cost to cut, waste the material, borrow and tip without overhaul will equal 2e. Economically e + Ic = 2e I = — (assuming no cost for wasting) Thus if the cost of excavation is 2/6 per yd 3 and the cost of over- haul is 2d per station yard, then the total economic overhaul distance = -y = 1500 ft If the free-haul is given as 500 ft the maximum economic haul = 1500 + 500 = 2000 ft. The overhaul distance is found from the mass-haul diagram by de- termining the distance from the centroid of the mass of the excavation to the centroid of the mass of the embankment. The centroid of the excavation and of the embankment can be 548 SURVEYING PROBLEMS AND SOLUTIONS determined (1) graphically, (2) by taking moments, (3) planimetrically. These methods are illustrated in the following example. Example 10.11 Volumes of cut and fill along a length of proposed road are as follows: Chainage Volume Cut (ft 3 ) Fill 100 290 200 760 300 1680 400 620 480 120 500 20 600 110 700 350 800 600 900 780 1000 690 1100 400 1200 120 Draw a mass diagram, and excluding the i surplus excavated mat- erial along this length determine the overhaul if the free-haul distance is 300 ft. (I.C.E.) Answer Chainage Volume (ft 3 ) A , Cut Fill Aggregate volume (ft 3 ) 100 290 + 290 JL\J\J 200 300 400 760 + 1050 1680 + 2730 620 + 3350 7UW 480 120 + 3470 500 600 20 + 3450 110 + 3340 700 350 + 2990 / \J\J 800 900 600 + 2390 780 + 1610 1000 690 + 920 1100 400 + 520 1200 3470 3070 120 3070 + 400 Check 400 VOLUMES 549 (ft 3 ) + 4 OCX) 3000 2000 1000 100 200f 300 400 500 600 700 800 |900 1000 1100 1200ft 240' 880' Fig. 10.45 Mass-haul diagram (1) Graphical Method (i) As the surplus of 400 ft 3 is to be neglected, the balancing line is drawn from the end of the mass-haul curve, parallel to the base line, to form a new balancing line ab. (ii) As the free-haul distance is 300 ft, this is drawn as a balanc- ing line cd. (iii) From c and d, draw ordinates cutting the new base line at (iv) To find the overhaul: (a) Bisect cc, to give c 2 and draw a line through c 2 parall- el to the base line and cutting the curve at e and /, which now represent the centroids of the masses acc % and dbd, (b) The average haul distance from acc^ in excavation to make up the embankment dbd x = ef. (c) The overhaul distance = the haul distance-the free-haul distance, i.e. ef - cd i.e. Scaled value = 640 - 300 = 340 ft. (d) The overhaul of material at acc x = volume (cc,) x overhaul distance (ef = 2750 ft 3 x 340 ft = 346' 3 station yards cd) (2) By taking moments With reference to the mass-haul curve and the tabulated volumes, moments are taken at a to find the centroid of the area acc^ . At a the chainage is scaled as 120 ft. 550 SURVEYING PROBLEMS AND SOLUTIONS Chainage (a) 120 - 200 200 - 300 300 - 350 (c) Volume (ft 3 ) 1050-400 = 650 2730- 1050 = 1680 3150-2730= 420 Iv = 2750 Thus the distance from a to the centroid 330500 Distance (ft) |(200-120) = 40 ^(300-200) + 80 = 130 | (350 -300)+ 180 = 205 Product (VxD) 26000 218400 86100 » 330 500 2750 = 120-2 ft /. Chainage of the centroid = 120 + 120-2 = 240*2 ft. Taking moments at d, chainage 650 ft: Chainage Volume (ft 3 ) Distance (ft) Product (VxD) (d) 650 - 700 3350-2990 = 160 J( 700- 650) = 25 4000 700- 800 2990-2390 = 600 |(800-700)+ 50 = 100 60 000 800- 900 2390-1610= 780 l(900-800) + 150 = 200 156 000 900-1000 1610- 920 = 690 |(1000 -900) + 250 = 300 207 000 1000-1100 920- 520 = 400 |(1 100 - 1000) + 350 =400 160000 1100-1200 - 520- 400 = 120 ^(1200- 1100) + 450 = 500 60 000 2v =2750 SP 647000 Thus the dists Mice from d to the < :entroid 647000 = 235* 3 ft 2750 Chainage of the centroid = 650 + 235*3 = 885*3 ft Average haul distance = 885*3 - 240*2 = 645*1 Length of overhaul = 645*1 - 300 = 345*1 rt . f 2750 x 345*1 oc1 c . .. Overhaul = — -— = 351*5 station 2700 yards. (3) Planimetric Method Distance to centroid = Haul/volume _ Area x horizontal scale x vertical scale volume ordinate From area acc^ Area scaled from mass-haul curve = 0*9375 in 2 Horizontal scale = 200 ft to 1 in. VOLUMES 551 Vertical scale = 1600 ft to 1 in. .-. Haul = 0-9375 x 200 x 1600 = 300000 Volume (ordinate cc,) = 2750 Distance to centroid = 300000/2750 = 109-1 ft Chainage of centroid = 350 - 109*1 = 240-9 ft For area dbd, Area scaled = 1-9688 in 2 .-. Haul = 1-9688 x 320000 = 630016 Volume = (ordinate dd,) = 2750 Distance to centroid = 229-1 ft Chainage of centroid = 650 + 229-1 = 879-1 ft Average haul distance = 879-1 - 240-9 = 638-2 ft Overhaul distance = 638-2 - 300 = 338-2 ft .-. Overhaul = 338*2 x 2750 = 344-5 station yards. N.B. Instead of the above calculation the overhaul can be obtained direct as the sum of the two mass-haul curve areas acc y and dbd, . . 300000 4 ' . Area acq = . station yd a aua 630016 „ 4 . Area aba^ = — jjnn station yd Total area = overhaul = . ftft = 344*5 station yards Proof Fig. 10.46 Take any area cut off by a balancing line, Fig. 10.46. Let a small increment of area 8 A = (say) 1 yd 3 and length of 552 SURVEYING PROBLEMS AND SOLUTIONS haul be /. 8A = 1 yd 3 x Z/100 station yd A = n x 1 yd x — n = Total volume x average haul distance, *. Area = Total Haul Exercises 10(c) (Earthwork volumes) 15. Calculate the cubic contents, using the prismoidal formula of the length of embankment of which the cross-sectional areas at 50 ft in- tervals are as follows: Distance (ft) 50 100 150 200 250 300 Area (ft 2 ) 110 425 640 726 1590 1790 2600 Make a similar calculation using the trapezoidal method and ex- plain why the results differ. (I.C.E. Ans. 11688 yd 3 ; 12085 yd 3 ) 16. The following notes were taken from the page of a level book: Reduced level Remarks 45-85 At peg 10 44-10 30 ft to right at peg 10 44-75 30 ft to left at peg 10 46-35 At peg 11 42-85 30 ft to right at peg 11 48-35 30 ft to left at peg 11 46-85 At peg 12 Draw cross-sections to a scale of 1 in. = 10 ft at pegs 10 and 11, which are 100 ft apart on the centre line of a proposed branch rail- way, and thereafter calculate the volume of material excavated between the two pegs in forming the railway cutting. The width at formation level is 15 ft, and the sides of the cutting slope at V/2 horizontal to 1 vertical. The formation level of each peg is 30*5 ft. (M.Q.B./M Ans. 2116 yd 3 ) 17. A level cutting is made on ground having a uniform cross-slope of 1 in 8. The formation width is 32 ft and the sides slope at 1 vertical to 1% horizontal. At 3 sections, spaced 66 ft apart, the depths to the centre line are 34, 28 and 20 ft. Calculate (a) the side widths of each section (b) the volume of the cutting. (N.R.C.T. Ans. 62-0; 96*6 ft; 53-3; 83-2 ft; 41*8, 65'2 ft; 11600 yd 3 ) VOLUMES 553 18. Calculate the volume in cubic feet contained between three suc- cessive sections of a railway cutting, 50 ft apart. The width of forma- tion is 10 ft, the sides slope 1 vertical to 2 horizontal and the heights at the top of the slopes in feet above formation level are as follows: Left Centre Right 1st Cross-section 13-6 12-0 14-0 2nd Cross-section 16-0 15-5 17-8 3rd Cross-section 18-3 16-0 16-0 (N.R.C.T. Ans. 63 670 ft 3 ) 19. The formation of a straight road was to be 40 ft wide with side slopes 1 vertically to 2^ horizontally in cutting. At a certain cross- section, the depth of excavation on the centre line was 10 ft and the cross-fall of the natural ground at right angles to the centre line was 1 vertically to 8 horizontally. At the next cross-section, 100 ft away, the depth on the centre line was 20 ft and the cross-fall similarly 1 in 10. Assuming that the top edge of each slope was a straight line, find the volume of excavation between the two sections by the prismoidal formula and find the percentage error that would be made by using the trapezoidal formula. (L.U. Ans. 11853 ft 3 12*5 %) 20. A straight embankment is made on ground having a uniform cross- slope of 1 in 8. The formation width of the embankment is 30 ft and the side slopes are 1 vertical to V/z horizontal. At three sections spaced 50 ft apart the heights of the bank at the centre of the formation level are 10, 15 and 18 ft. Calculate the volume of the embankment and tabulate data required in the field for setting out purposes. (L.U. Ans. 2980 yd 3 ) 21. Cross-sections at 100 ft intervals along the centre line of a pro- posed straight cutting are levelled at 20 ft intervals from -60 ft to + 60 ft and the following information obtained: Distances (ft) -60 -40 -20 +20 +40 +60 4-0 1-0 0-0 0-0 0-0 1-0 2-8 100 12-9 8-6 5-0 3-0 2-0 3*0 6-0 200 17-5 14-1 10-9 8-0 6-0 6-0 9-6 300 21-8 17-7 14-4 11-3 9'7 9*7 11-0 400 25-0 21-2 18-0 15*2 12*8 12-0 13*2 (Tabulated figures are levels in feet relative to local datum). The for- mation level is zero feet, its breadth 20 ft, and the side slopes 1 ver- tical to 2 horizontal. Find the volume of excavation in cubic yards over the section given. (L D Ans 510 o y d 3 ) 554 SURVEYING PROBLEMS AND SOLUTIONS 22. A minor road with a formation width of 15 ft is to be made up a plane slope of 1 in 10 so that it rises at 1 in 40. There is to be no cut or fill on the centre line, and the side slopes are to be 1 vertical to 2 horizontal. Calculate the volume of excavation per 100 ft of road. Derive formulae for calculating the side-widths and heights and the cross-sectional area of a 'two-level' section. (N.R.C.T. Ans. 25 yd 3 /100 ft) 23. The uniform slope of a hillside (which may be treated as a plane surface) was 1 vertically to 4 horizontally. On this surface a straight centre line AB was laid out with a uniform slope of 1 vertically to 9 horizontally. With AB as the centre line a path with a formation width of 10 ft was constructed with side slopes of 1 vertically to 2 horizontally. If the path was 500 ft in length and there was no cut or fill on the centre line, calculate the quantity of cutting in cubic feet. (I.C.E Ans. 2530 ft 3 ) 24. The central heights of the ground above formation at three sections 100 ft apart are 10, 12 and 15 ft and the cross-falls at these sections 1 in 30, 1 in 40 and 1 in 20 (vertically to horizontally). If the forma- tion width is 40 ft and the side slopes 1 vertically in 2 horizontally, calculate the volume of excavation in the 200 ft length (a) if the centre line is straight, (b) if the centre line is an arc of 400 ft radius. (L.U. Ans. 158 270 ft 3 ; 158 270 ±1068 ft 3 ) 25. The centre line of a highway cutting is on a curve of 400 ft radius, the original surface of the ground being approximately level. The cut- ting is to be widened by increasing the formation width from 20 to 30 ft, the excavation to be entirely on the inside of the curve and to retain the existing side slopes of 1V 2 horizontal to 1 vertical. If the depth of formation increases uniformly from 8 ft at ch. 600 to 17 ft at ch. 900, calculate the volume of earth to be removed in this 300 ft length. (L.U. Ans. 1302 yd 3 ) 26. The contoured plan of a lake is planimetered and the following values obtained for the areas enclosed by the given underwater con- tours: Contour (ft O.D.) 305 300 295 290 285 Area (ft 2 ) 38500 34700 26 200 7800 4900 The surface area of the water in the lake is 40 200 ft 2 . The top water level and the lowest point in the lake are at 308-6 and 280*3 ft O.D. respectively. Find the quantity of water in the lake in millions of gallons. (L.U. Ans. 3-73 m. gal) VOLUMES 555 27. The areas of ground within contour lines at the site of a reservoir are as follows: Contour in ft above datum Area (ft 2 ) 400 505602 395 442 104 390 301 635 385 232203 380 94056 375 56821 370 34 107 365 15834 360 472 Taking 360 ft O.D. as the level of the bottom of the reservoir and 400 ft O.D. as the water level, estimate the quantity of water in gallons contained in the reservoir (assume 6*24 gal per ft 3 ). (Ans. 45 276500 gal) 28. Describe three methods of carrying out the field work for obtaining the volumes of earthworks. Explain the conditions under which the 'end area' and 'prismoidal rule' methods of calculating volumes are accurate, and explain also the use of the 'prismoidal correction'. The areas within the contour lines at the site of a reservoir are as follows: Contour (ft) Area (ft 2 ) 400 5 120 000 395 4642000 390 4060000 385 3184000 380 2356000 375 1 765 000 370 900000 365 106000 360 11000 The level of the bottom of the reservoir is 360 ft. Calculate (a) the volume of water in the reservoir when the water level is 400 ft using the end area method, (b) the volume of water in the reservoir using the prismoidal formula (every second area may be taken as a mid- area), and (c) the water level when the reservoir contains 300000000 gallons. (L.U. Ans. (a) 97-8925 m. ft 3 (b) 97-585 m. ft 3 388 ft) 29. A square level area ABCD (in clockwise order) of 100 ft side is to be formed in a hillside which is considered to have a plane surface with a maximum gradient of 3 (horizontally) to 1 (vertically). 556 SURVEYING PROBLEMS AND SOLUTIONS £ is a point which bisects the side AD, and the area ABE is to be formed by excavation into the hillside, whilst the area BCDE is to be formed on fill. The side slopes in both excavation and fill are to be 1 to 1, and adjacent side slopes meet in a straight line. By means of contours at 2 ft intervals, plot the plan of the earth- works on graph paper to a scale of 50 ft to 1 inch. Hence compute the volume of excavation. (I.C.E. Ans. V ~ 880 yd 3 ) 30. A road having a formation width of 40 ft with side slopes of 1 in 1 is to be constructed. Details of two cross- sections of a cutting are as follows: Depth of Cutting Side Slope Limits (ft) Chainage (it) on Centre Line (ft) Left Right 500 10-2 25-2 33-7 600 6-0 22-0 28-5 Assuming that these cross-sections are bounded by straight lines and that the undisturbed ground varies uniformly between them, com- pute the volume of excavation allowing for prismoidal excess. If instead of being straight, the plan of the centre line had been a circular curve of radius R with the centre of curvature on the right, how would this have been taken into account in the foregoing calcula- tions? Quote any formula that would have been used. (I.C.E. Ans. 1370 yd 3 ) 31. A section of a proposed road is to run through a cutting from chain- age 500 to 900, the formation level falling at 1 in 200 from chainage 500. The formation width is to be 30 ft and the side slopes are to be 1 vertical to 2 horizontal. The original ground surface is inclined uniformly at right-angles to the centre line at an inclination of 1 in 10. With the information given below, calculate the volume of excava- tion in cubic yards, using the prismoidal formula. ->, . ^ .. T , Ground Level at Chainage Formation Level « ^ T • Centre Line 500 44-25 ft 51-11 600 50-82 700 50-93 800 51-09 900 50-77 (I.C.E. Ans. 5474 yd 3 ) 32. A road of 40 ft formation width is to be constructed with side slopes of 1 (vertical) to V-/i (horizontal) in excavation and 1 (vertical) to 2 (horizontal) in fill. Further details of two cross-sections are given VOLUMES 557 below where the cross fall of the undisturbed ground is 1 (vertical) to r (horizontal). Chainage (ft) Ground level on Formation Level Centre-line (ft above datum) (ft above datum) 400 171-6 166-6 4 500 170-2 168-0 6 Assuming the road is straight between these two sections, compute the volumes of excavation and fill in 100 ft length neglecting prismoi- dal excess. (I.C.E. Ans. 819 yd 3 cut, 11 yd 3 fill) 33. On a 1000 ft length of new road the earthwork volumes between sections at 100 ft intervals are as follows, the excavation being taken as positive and filling as negative: Section No 01234 56 7 8 9 10 Vol. (1000 yd 3 ) 3-7 9-1 15-0 13-9 6-4 1-4 -5-6 -19-4 -18-9-5-6 Draw the mass- haul curve and find (i) the volume to be moved under the terms of the free-haul limit of 300 ft, (ii) the volume to be moved in addition to (i), (iii) the number of station-yards under (ii) where 1 station-yard equals 1 cubic yard moved 100 ft, (iv) the average length of haul under (ii). (L.U. Ans. 7500 yd 3 ; 42 000 yd 3 ; 23500 station yds; 560 ft) 34. The following figures show the excavation (+) and filling (-) in cubic yards between successive stations 100 ft apart in a proposed road. 1 2 3 4 5 6 + 1500 + 1100 + 500 + 100 - 100 -1000 7 8 9 10 11 12 -2200 -2500 -1600 -400 +1800 + 2800 State which of the following tenders is the lower and calculate the total mass haul in the 1200 ft length: (a) Excavate, cart and fill at 9/6 per yd 3 . (b) Excavate, cart and fill at 9/- per yd 3 , with a free-haul limit of 400 ft, plus 1/- per station yard for hauling in excess of 400 ft. (1 station yard = 100 ft x 1 yd 3 ). (L.U. Ans. 28 500 station yards; (a) £3700, (b) £3750) 35. Volumes in yd 3 of excavation (positive) and fill (negative) between successive sections 100 ft apart on a 1300 ft length of a proposed railway are given in the following table: 558 SURVEYING PROBLEMS AND SOLUTIONS Section 1 2 3 4 5 6 Volume -1000 -2200 -1600 -500 +200 +1300 7 8 9 10 11 12 13 + 2100 +1800 +1100 +300-400 -1200 -1900 Draw a mass haul curve for this length. If earth may be borrowed at either end, which alternative would give the least haul ? Show on the diagram the forward and backward free-hauls if the free-haul limit is 500 ft, and give these volumes. (L.U. Ans. Borrow at end, 1150 ft; 2900 yd 3 ; 2400 yd 3 ) 36. The volumes in yd 3 between successive sections 100 ft apart on a 900 ft length of a proposed road are given below; excavation is shown positive and fill negative Section 012 3 4 56 78 9 Volume +1700 -100 -3200 -3400 -1400 +100 +2600 +4600+1100 Determine the maximum haul distance when earth may be wasted only at the 900 ft chainage end. Show and evaluate on your diagram the overhaul if the free-haul limit is 300 ft. (L.U. Ans. 510 ft; 4950 station yards) Bibliography bannister, a. and RAYMOND, s., Surveying (Pitman) DAVIES, R.E. FOOTE, F.S., and KELLY, J.W., Surveying Theory and Practice (McGraw Hill) CLARK, D., Plane and Geodetic Surveying, Vol.1 (Constable) THOMAS, W.N., Surveying (Edward Arnold) CURTIN, w. and LANE, R.F., Concise Practical Surveying (English Universities Press) RITTER-PAQUETTE, Highway Engineering, 2nd Ed. (Ronald Press) SHARMA, R.c. and SHARMA, S.K. Principles and Practice of Highway Engineering (Asia Publishing House) MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.) 11 CIRCULAR CURVES 11.1 Definition The curve can be defined by (a) the radius R or (b) the degree of the curve D. D can be expressed as the angle at the centre of the curve subtended by (i) a chord of 100 ft or (ii) an arc of 100 ft (the former is more generally adopted). In Fig. 11.1, (11.1) . , _ i x 100 50 Sm ^ " 2 R " R r R- 50 sin^D D is small, sin | D = % D radians 206265x50 K = ~ 5730 \D° x3600 D° (11.2) 100' Fig. 11.1 11.2 Through Chainage Through chainage represents the length of road or rail from some terminus (it does not necessarily imply Gunter's chain — it may be the engineer's chain). Pegs are placed at 'stations' (frequently at 100 ft intervals) and a point on the construction can be defined by reference to the * station'. When a curve is introduced, Fig. 11.2, the tangent point T, is said to be of chainage 46 + 25, i.e. 4625 ft from the origin. If the length of the curve was 400 ft, the chainage of T z would be expressed as (46 + 25) + (4 + 00), i.e. 50 + 25. Origin of system 12 3 T, 46+25 42 43 44 45 46/47 Chainage of 7i — 46+25 Length of arc » 400ft Chainage of T, ■ 50+25 Fig. 11.2 Through chainage 559 560 SURVEYING PROBLEMS AND SOLUTIONS 11.3 Length of Curve (L) L = 27TR * §fo (11 - 3) or L = R.6 Tad (11.4) The second formula is better. Example 11.1 If 6 = 30° 26 and R = 100 ft, 26 2 x 3-142 x 100 x 30^ 60 Length of arc L = — — = 53 '12 ft or L = 100 x 30°26; ad - 100 x 0*53116 = 53-12 ft 11.4 Geometry of the Curve In Fig. 11.3, T, and T 2 are tangent points, / is the intersection point, is the deflection angle at inter- section point, /T, = IT 2 - tangent length = Rtan±<j>. T, T 2 = Long chord = 2 . TX = 2Ksin^c£ (11.5) T,A = chord c = 2R sin a (11.6) (if c < R/20 and a is small, arc ~ chord, i.e. sin a = a rad ) Fig. 11.3 Geometry of the curve c = 2Ra Tad (11.7) Deflection angle a rad = S- (11.8) 206 265 c a sec= — 2£ (119) 206265c a — = 2^60" (U - 10) . l-Z^£ (ii.iD CIRCULAR CURVES 561 10 R sec \ <f> (11.12) IP = 10 - PO = R sec±<f> - R = R(sec\cf> - 1) (11.13) PX = PO - XO = R - flcosl0 = /?(1- cos^) (11.14) = R versine^0. (11.15) 11.5 Special Problems 11.51 To pass a curve tangential to three given straights (Fig. 11.4) (1) + (2) i.e R = T y Xcot±j> = (^ + z)cot|0 = (* 2 + y)cot^<£ = {{(x + z-y) + y)cot^<£ = ^(x + y + z) cot^<£ R = 5cot^0 where s = \ perimeter of A XYZ (11.16) 562 SURVEYING PROBLEMS AND SOLUTIONS Alternative solutions: (a) Area XOY = \Rz XOZ = ±Ry YOZ = ±Rx XYZ = areas (XOY + XOZ - YOZ) = \R(y+z-x) = ±R(x + y+z)- \R(x+ x) = Rs - Rx = R(s - x) area A XYZ :. R = . (11.17) S - X (b) If angles a and /3 are known or computed, YP = R tan a/2 PZ = R tan 0/2 .-. YZ = YP + PZ = K(tan a/2 + tan /S/2) YZ R = tana/2 + tan /S/2 (1118) Example 11.2 The co-ordinates of three stations A, B and C are as follows :- A E 1263-13 m N1573'12m B E 923*47 m N 587'45m C E 1639-28 m N 722-87 m. The lines AB and AC are to be produced and a curve set out so that the curve will be tangential to AB, BC and AC. Calculate the radius of the curve. In Fig. 11.5, ,923-47- 1263-13 Bearing AB = tan" 587-45 - 1573-12 , -, -339-66 = tan ' - 985-67 = S19°00' 50" W = 199° 00' 50". Fig. 11.5 CIRCULAR CURVES 563 Length AB = 985*67 sec 19° 00' 50" = 1042-55 n- Q •„ An . _,1639-28- 1263*13 _, 376*15 Bearing AC = tan ' = tan ' 722-87 - 1573-12 - 850-25 = S23°51'52"E = 156°08'08". Length AC = 850*25 sec 23° 51' 52" = 929*74 Bearing EC - tan -, 1639*28 - 923*47 = ^-,715*81 722-87 - 587-45 135-42 = N79°17'14"E = 079° 17' 14" Length BC = 135*42 sec 79° 17' 14" = 728*51 = 180 - (19° 00' 50" + 23° 51' 52") = 137°07'18" To find the radius Using Eq. (11.16), R = {(AB+BC+ AC) cot \4> = ^(1042*55 + 728*51 + 929*74) cot 68° 33' 39" = 1350*40 cot 68° 33' 39" = 530*2 m. Alternatively, by Eq. (11.17), R = V / fc( 5 - a >( s - b H s - c >i s — a = Vl 1350*40(1350*40 - 728*51)(1350*40 - 929*74) (1350*40 - 1042*54)} 1350*40 - 728*51 = 530*2 m. Alternatively, by Eq. (11.18), a = 079° 17' 14" - 019° 00' 50" = 60° 16' 24" )8 = 336° 08' 08" - 259° 17' 14" = 76° 50' 54" check <f> = 137° 07' 18" o = 728*51 tan 30° 08' 12"+ tan38°25'27" " ^L£B^ 11.52 To pass a curve through three points (Fig. 11.6) In Fig. 11.6, angle COA = 20 angle XOC = 180 - XC = ±AC = R sin (180-0) 564 SURVEYING PROBLEMS AND SOLUTIONS AC sin i = 2R, \AC i.e. R = (11.19) sin 6 Similarly, the full sine rule is written AB sinC BC sin A AC sin B = 2R (11.20) Example 11.3 The co-ordinates of two points B and C with respect to A are B 536*23 mN 449*95 m E C 692-34 mN 1336'28mE. Calculate the radius of the circular curve passing through the three points. 449*95 Bearing AB = tan" 1 ——— = N40°E boo'Zo Bearing BC = tan^ fff'ff = N80°E 156*11 .-. angle ABC = 6 = 180 + 40 - 80 = 140°. Length AC = V(692*34 2 + 1336-28 2 ) = 1505 m 1505 752*5 _ 2 sin 140° " sin 40 " t±I}LL Example 11 .4 In order to find the radius of an existing road curve, three suitable points A, B, and C were selected on the centre line. The instrument was set up at B and the following tacheometrical read- ings taken on A and C, the telescope being horizontal and the staff held vertical in each case. Staff at Horizontal angle Collimation Stadia A 0°00' 4*03 5*39/2*67 C 195° 34' 6*42 8*04/4*80. CIRCULAR CURVES 565 If the instrument had a constant multiplier of 100 and an additive constant of zero, calculate the radius of a circular arc ABC. If the trunnion axis was 5*12 ft above the road at B, find the gradients of AB and BC. (L.U.) In Fig. 11.7, AB = 100(5-39-2-67) = 272 ft BC = 100(8-04-4-80) = 324 ft. In triangle ABC, . A-B 324- 272 „ 195°34'-180 tan — - — = tan 2 324 + 272 2 52 tan7°47' 596 Fig. 11.7 A-B R = 2 V tJ. A + B 2 7° 47' A = 8° 28' B = 7° 06' BC sin A 2R 324 1 1 fifi»3 ff 2 sin 8° 28' " 11003ft - Difference in height A-B, 5*12 - 4*03 = 1*09 ft. Gradient AB, 1*09 in 272 = 1 in 240. Difference in height B - C, 6'42 - 5*12 = 1*30 ft. Gradient BC, 1 -30 in 324 = 1 in 249. If the gradients are along the arc, = 2 sin -1 324 = 16° 56' 2 x 1100-3 Length of arc AB = 1100'3 x 16° 56 ' rad - 272 '7 ft 6> 2 = (2x15° 34')- 16° 56' = 14° 12' Length of arc BC = 1100-3 x 14° 12; ad = 352'2 ft. Gradients along the arc are 1 in 251; 1 in 250. 566 SURVEYING PROBLEMS AND SOLUTIONS Exercises 11 (a) 1. It is required to range a simple curve which will be tangential to three straight lines YX, PQ, and XZ, where PQ is a straight, joining the two intersecting lines YX and XZ. Angles YPQ = 134° 50' ; YXZ = 72° 30' ; PQZ = 117° 40" and the distance XP = 5'75 chains. Compute the tangent distance from X along the straight YX and the radius of curvature. (I.C.E.. Ans. 8*273 chains; 6*066 chains) 2. A circular road has to be laid out so that it shall be tangent to each of the lines DA, AB, and BC. Given the co-ordinates and bearings as follows, calculate the radius of the circle. Latitude Departure A - 29*34 m - 128*76 m B - 177-97 m - 58-39 m Bearing DA 114° 58' 10" CB 054° 24' 10" (Ans. 137-48 m) 3. A circular curve is tangential to three straight lines AB, BC, and CD, the whole circle bearings of which are 38°, 72° and 114° respec- tively. The length of BC is 630 ft. Find the radius and length of the curve and the distances required to locate the tangent points. Also tabulate the data necessary for setting out the curve from the tangent point on BC with 100 ft chords and a theodolite reading to 30" with clockwise graduations. (L.U. Ans. K = 913-6ft; 1211'9ft; 279-3 and 350'7ft) 4. The co-ordinates of two stations Y and Z in relation to a station X are as follows: E(m) N(m) Y 215 576 Z 800 -750 Find by construction the radius of a circular curve passing through each of the stations X, Y, and Z . (Ans. 790 m) 5. Three points A, B, and C lie on the centre line of an existing mine roadway . A theodolite is set up at B and the following observa- tions were taken on to a vertical staff. Staff at Horizontal circle Vertical circle c . ,. ~ ... Stadia Collimation A 002° 10' 20" +2° 10' 6*83/4-43 5*63 C 135° 24' 40" -1°24' 7-46/4-12 5'79 CIRCULAR CURVES 567 If the multiplying constant is 100 and the additive constant zero, calculate: (a) the radius of the circular curve which will pass through A, B and C, (b) the gradient of the track laid from A to C if the instrument height is 5-16 ft. (R.I.C.S./M Ans. 362-2 ft; 1 in 33-93) 11.53 To pass a curve through a given point P (a) Given the co-ordinates of P relative to I, i.e. I A and AP (Fig. 11.8) I. ♦/ 2) VL N 0* Jfl) \r JT^^^Sv \ X X \ \ N: / f * / / / / t Fig. 11.8 In Fig. 11.8, Assume BC passing through P is parallel to 7, T z . AB = AP cot \<f> BP = AP cosec \<f> = QC BX = (AB + AI) cos \<f> PX = BX - BP BQ = BX + PX T,B = y/(BP x BQ) (intersecting tangent and T,/ = BI ±T,B (i.e. T, 1 ^ = T, B) R = T x l cot - <f> or T, 1 / cot \<f> secant) = (AB + AI ±T,B) cot \<f> (11.21) 568 SURVEYING PROBLEMS AND SOLUTIONS (b) Given polar co-ordinates of point P from intersection I (Fig. 11. 9) In Fig. 11. 9, a = 90-(f +0) 01 = R sec|<£ 01 _ Rsec\cf> OP R ' also by sine rule sin (a + jS) sin a ••• sin (a + /3) = sin a sec ^<f> [ = sinl80-(a + i 8)] (11.22) Thus there are 2 values for /3. R IP sin a sin /3 R = IP sin a cosec j8 Fig. 11.9 (2 values) (11.23) Fig. 11.10 CIRCULAR CURVES 569 If a = 0, in Fig. 11.9, i.e. P lies on line 10, 1 01 = R sec 10 = IP ± R R(sec\c/> +1)= IP R = IP secl<£ ± 1 If 0=90°, in Fig. 11.10, T,0 = R QP =!,/ = /? sinS - R tan^c£ sin 8 = tan^<£ 7,(3 = IP = R - RcosS = K(l -cosS) verso As 5 has 2 values there are two values of R . (11.24) (11.25) Example 11.5 A circular railway curve has to be set out to pass through a point which is 40 m from the intersection of the straights and equidistant from the tangent points. The straights are deflected through 46° 40 ' . Calculate the radius of the curve and the tangent length. Taking the maximum radius only, in Fig. 11.11, 01 = OP + PI = R + x = R sec^<£ .*. R(l - sec|0)= - x i.e., Eq. (11.24), x R 40 secl<£ - l 1-089 07- 1 449-1 m. Fig. 11.11 TJ = T 2 I = R tan-^0 = 449-1 tan 23° 20' 193-7 m. 570 SURVEYING PROBLEMS AND SOLUTIONS Example 11.6 The bearings of two lines AB and BC are 036° 36' and 080° 00' respectively. At a distance of 276 metres from B towards A and 88 metres at right-angles to the line AB, a station P has been located. Find the radius of the curve to pass through the point P and also touch the two lines. (R.I.C.S./M) Taking the maximum radius only In Fig. 11.12, Fig. 11.12 cf> = 080° 00' - 036° 36' = 43° 24' In triangle GLP, LG = LP cot \<$> = 88cot21°42' = 221*13 m GP = LPcosec±(f> = 88cosec21°42' = 238*00 m GX = (LG + LB) cos^<£ = (221*13 + 276*00) cos 21° 42' = 461*90m PX = GX - GP = 461*90 - 238*00 = 223*90 m GQ = GX + PX = 461*90 + 223*90 = 685*80 m T,G= y/(GP x GQ) = V(238*00 x 685*80) = 404*01 m T,B = GB + T,G = BL + LG + GT, = 276*0 + 221*13 + 404*01 = 901*14 m R = T,B cot|0 = 901*14 cot 21° 42' = 2264*5 m. CIRCULAR CURVES 571 Alternative solution 6 = tan"' 88/276 = 17° 41' a = 90 - (±<f> + 6) = 90 - (21° 42' + 17° 41') = 50° 37' x = 276 sec0 = 289*69 m Then, from Eq. (11.22), sin(a+/8) = sinasec^0 = sin 50° 37 'sec 21° 42' a+/3 = 56° 17' 30" a = 50° 37' 00" .-. B = 5° 40 '30" Therefore radius R = x sin a cosec j8 = 289-69 sin 50° 37' cosec 5° 40' 30" = 2264*5 m Exercises 11. (b) (Curves passing through a given point) (N.B. In each case only the maximum radius is used) 6. In setting out a circular railway curve it is found that the curve must pass through a point 50 ft from the intersection point and equi- distant from the tangents. The chainage of the intersection point is 280 + 80 and the intersection angle (i.e. deflection angle) is 28°. Calculate the radius of the curve, the chainage at the beginning and end of the curve, and the degree of curvature . (I.C.E. Ans. 1633 ft; 276 + 73; 284 + 73; 3° 30') 7. Two straights intersecting at a point B have the following bear- ings: BA 270°; BC 110°. They are to be joined by a circular curve, but the curve must pass through a point D which is 150 ft from B and the bearing of BD is 260°. Find the required radius, the tangent distances, the length of the curve and the deflection angle for a 100 ft chord. (L.U. Ans. 3127*4 ft; 551*4 ft; 1091*7 ft; 0°55') 8. The co-ordinates of the intersection point / of two railway straights, AI and IB, are 0,0. The bearing of AI is 90° and that of IB is 57° 14'. If a circular curve is to connect these straights and if this curve must pass through the point whose co-ordinates are - 303*1 ft E and + 20*4 ft N, find the radius of the curve. 572 SURVEYING PROBLEMS AND SOLUTIONS Calculate also the co-ordinates of the tangent point on Al and the deflection angles necessary for setting out 100 ft chords from this tangent point. What would be the deflection angle to the other tangent point and what would be the final chord length? (L.U. Ans. 2000ft; 588*0 ft E Oft N) 9. A straight BC deflects 24° right from a straight AB. These are to be joined by a circular curve which passes through a point D 200 ft from from B and 50 ft from AB. Calculate the tangent length, the length of the curve, and the def- lection angle for a 100 ft chord. (L.U. Ans. 818-6; 806*5; 0° 45') 10. A right-hand circular curve is to connect two straights Al and IB, the bearings of which are 70° 42' and 130° 54' respectively. The curve is to pass through a point X such that IX is 132*4 ft and the angle AIX is 34° 36'. Determine the radius of the curve. If the chainage of the intersection point is 5261 ft, determine the tangential angles required to set out the first two pegs on the curve at through chainages of 50 ft. (L.U. Ans. 754*1 ft; 0° 59' 40"; 2° 53' 40") 11. The following readings were taken by a theodolite stationed at the point of intersection / of a circular curve of which A and B are re- spectively the first and second tangent points A 14° 52'; B 224° 52'; C 344° 52' It is required that the curve shall pass through the point C, which is near the middle of the curve, at a distance of 60 ft from /. (a) Determine the radius of the curve. (b) Calculate the running distances of the tangent points A and B and the point C, the distance at / being 200 + 72, in 100 ft units. (c) Show in tabular form the running distances and tangential angles for setting out the curve between A and C. (L.U. Ans. 1181ft; 197 + 56; 203 + 74; 200 + 22) 11.54 Given a curve joining two tangents, to find the change required in the radius for an assumed change in the tangent length (Fig. 11.13) In Fig. 11.13, R 2 -R, = (*a-',)cotJL0 (11.26) o,o 2 = (R 2 - /?,) secj0 (11.27) CIRCULAR CURVES 573 A, A, — R, = IX 2 - IX, (/0 2 - R 2 ) - (70, - /?,) (R 2 sec {$ - R 2 ) - (K, sec \<f> - /?,) /?2(sec|0 - 1) - R,(sec\cf> - 1) (/? 2 -/?,)(secl0-l) X,X 2 sec-i0- 1 + «i (11.28) (11.29) Fig. 11.13 Example 11.7 (a) A circular curve of 2000 ft radius joins two points A and C which lie on the two straights AB and BC. If the running chainage values of A and C are 1091ft and 2895 ft respectively, calculate the distance of the midpoint of the curve from B. (b) If the minimum clearance value of the curve from B is to be 200ft, what radius would be required for the curve and what would be the chainage value for the new tangent points? (R.I.C.S.) (a) In Fig. 11.14, Length of curve - 2895 - 1091 = 1804 ft = R<k rad 0. rad — arc ~R 574 SURVEYING PROBLEMS AND SOLUTIONS - 1804 " 2ooo <f> = 51° 41' In triangle T^BO u BO, = R sec^0 = 2000sec25°50'30" = 2222-22 ft BX, = 2222-22 - 2000 = 222-22 ft In triangle T 3 BO z , By Eq. (11.29), 8/51*41' 1187-86 r , j 2811-46 Fig. 11.14 /?, = R s X,X 2 sec -^ <f> - 1 - 2000 -•• 222-22 - 200 1-1111-1 - 2000 - 200 = 1800 ft (b) By Eq. (11.27), 0,0 2 = (R z - /?,) sec ±cf> = (2000 - 1800) sec 25° 50' 30" = 222-22 ft TJ 3 = 2 P = 0,0 2 sin\c/> = 222*22 sin 25° 50 '30" = 96*86 ft Alternatively, by Eq. (11.26), e 2 - ',) = ( R z " *,) tan I <f> = 200 tan 25° 50' 30" = 96-86 ft Chainage T 3 = T, + (t 2 _ ^) = 1091 + 96*86 = 1187*86 ft CIRCULAR CURVES 575 Length of arc = 1800 x <f> = 1800 x 0*902 = 1623-60 .'. Chainage T A = 1187-86 + 1623*60 = 281 1-46 ft 11.6 Location of Tangents and Curve // no part of the curve or straights exists, setting out is related to a development plan controlled by traverse and/or topographical detail . // straights exist (1) Locate intersection point /. (2) Measure deflection angle <£. (3) Compute tangent length if radius R is known . (4) Set off tangent points T, and T z . // the intersection point is inaccessible Select stations A and B on the straights, Fig. 11.15. Measure a and /3; AB. Solve triangle AIB. AI = AB sin /8 cosec <f> BI = AB sin a cosec <f> TJ = T 2 I = R tan|0 /. T,A = T,/ - AI T 2 B = T,/ - Bl Fig. 11.15 For through chainage Chainage of A known Chainage of / = Chainage A + AI Chainage of T, = Chainage / - T,/ Chainage of T z = Chainage T, + arc T, T 2 576 SURVEYING PROBLEMS AND SOLUTIONS // the tangent point is inaccessible If T, is not accessible set out where possible from T 2 , Fig. 11.16. A check is possible by selecting station B on the curve and checking offset AB. Fig. 11.16 N.B. To locate the curve elements a traverse may be required joining the straights. It is essential to leave permanent stations as reference points for ultimate setting out. 11 .7 Setting out of Curves 11 .71 By linear equipment only (a) Offsets pom the long chord (Fig. 11.17) If, in Fig. 11.17, the chord is sub divided into an even number of equal parts, the offsets \h z etc. can be set out. Each side of the mid- point will be symmetrical. Generally, (R - yf = R z - x z y = R - y/(R 2 - x z ) (11.30) y 2 = R - y/(R z -xl) (11.31) .-. K - y - y 2 (11.32) N.B. If <f> is known, y = R(l - cosi<£), i.e. y = R versine^. Example 11.8 A kerb is part of a 100ft radius curve. If the chord joining the tangent points is 60ft long, calculate the offsets from the chord at 10ft intervals, Fig. 11.18. CIRCULAR CURVES 577 Fig. 11.18 As above (Eq. 11.30), h 3 = y = R -y/(R 2 -x 2 ) = 100 - V(100 2 -30 2 ) = 100 - V(100 - 30) (100 + 30) = 4-61 . By Eqs.11.31/11.32, h 2 = h 3 - y 2 y z = 100 - V(100 2 - 10 2 ) = 0-50 h 2 = 4*61 - 0*50 - 4J1 ft, = h 3 - y y y, = 100 - V(100 2 -20 2 ) = 2-02 ft, = 4-61 - 2-02 = 2-59 . (b) Offsets from the straight (Fig. 11.19) Fig. 11.19 As above, i.e., by Eq. (11.30), Alternatively, Fig. 11.20 (R - xf = R z - y 2 x = R - y/(R 2 -y 2 ) sin a = If a is small, c ~2±y. 2R 2R 2R (11.33) (11.34) (11.35) 578 SURVEYING PROBLEMS AND SOLUTIONS (c) Offsets from the bisection of the chord (Fig. 11.20). From Eq. (11.30), AA< _ R _^ [Rl _(c^ x Alternatively, AA, = R(l -cos-^) BB, = K(l -cos{<£) CC, = RQ. -cos^) (d) Offsets from the bisection of successive chords; centre of curve fixed (Fig. 11.21). As above, sin a c 2R Fig. 11.21 Assuming equal chords, Tt = Aa = Bb = Dd etc. = K(l - cos a) = R vers a Alternatively, Tt = R -**'-(§)" } Lay off T,x = c 2 Lay off xA = Tt Aa = Tt along line AO T, B on line Ta produced Bb = Tt along line BO (e) Offsets from chords produced. (i) Equal chords (Fig. 11.22) A Z A = c sin a = C X 2R 2R (Eq. 11.34) CIRCULAR CURVES 579 If a is small, A 2 A ^ A X A. This can be controlled if the length of chord is limited, i.e. R C< 20 B,B = 2A X A if a is small. •»2 B,B = R (11.36) (ii) Unequal chords (through chainage) Fig. 11.23 In Fig. 11.23, Offset from sub-chord X = A X A = —^ Offset from 1st full chord 2 = B,B = B X B 2 + B 2 B (Eq. 11.34) C 1 C 2 f|_ C 2 ( C 1 + C 2 ) 2/2 + 2/2 " 2/2 (11.37) 580 SURVEYING PROBLEMS AND SOLUTIONS Offset from 2nd full chord = 3 = D, D = D y D 2 + DJ) = 2D,D = 2B.B By Eq. (11.37), Q, = c 3 ( c 2 + c 3 ) 2K (11.38) but c 3 = c 2 Generally, a = 2/? Cn(Cn + C n -,) 2i? (11.39) 11.72 By linear and angular equipment (a) Tangential deflection angles (Fig. 11.24) By Eq. (11.33), sin a c = 2R If a is small ( c < R/20) sin a ~ a rad .-. 206 265 c a sec ~ 2R (11.40) 206 265 c 1718-8 c 2R x 60 R (11.41) Fig. 11.24 4> For equal chords 2 a = — (11.42) n where n = no. of chords required, For sub-chords, sub- chord a. ax standard chord (11.43) Fig. 11.25 (b) Deflection angles from chord produced (Fig. 11.25) The theodolite is successively moved round the curve. This method is applicable where sights from T, are restricted. It is the method applied underground. 11.73 By angular equipment only (Fig. 11.26) Deflection angles are set out from each tangent point, e.g., A is the intersection of a from T, with 3 a from T 2 . CIRCULAR CURVES N.B. na = ±<f> Generally, if the deflection angle from T. is a. 581 (11.44) the deflection angle jS from T 2 = 360 - l<f> + a, (11.45) Fig. 11.26 Example 11.9 In a town planning scheme, a road 30 ft wide is to in- tersect another road 40 ft wide at 60°, both being straight. The kerbs forming the acute angle are to be joined by a circular curve of 100 ft radius and those forming the obtuse angle by one of 400 ft radius. Calculate the distances required for setting out the four tangent points. Describe how to set out the larger curve by the deflection angle method and tabulate the angles for 50 ft chords. (L.U.) b d *■ c \ 1' Fig. 11.27 7, X = T Z X = 400 tan 30 = 230-94 ft T 3 Z = I 4 Z = 100 tan 60 = 173-21 ft In triangle XYb, 20 XY = YZ = 23-09 sin 60 Xb = 20 tan 30 = Yc = 11-55 582 SURVEYING PROBLEMS AND SOLUTIONS In triangle BYa, 15 BY = — = 17-33 sin 60 aY = Bd = 15 tan 30° = 8-66 Distances to tangent points measured along centre lines: AB = T X X+XY - aY = 230-94 + 23-09 - 8-66 = 245-37 ft BC = T A Z + XY + aY » 173-21 + 23-09+ 8-66 = 204-96 ft BD = T Z X -Xb +BY = 230-94 - 11-55 + 17-33 = 236-72 ft BE = T 3 Z + Xb + BY = 173-21 + 11-55 + 17-33 = 202-09 ft Deflection angles, 50 ft chords 50 sina i = « 7^ 1 2 x 400 a, = 3°35' By approximation, 206 265x50 __ a_ = -— — : = 12892 sec 2x400 = 3°34'52" a, = 3°35' a 2 = 7°10' a 3 = 10°45' a 4 = 14°20' a 8 = 17°55' a 6 = 21°30' a 7 = 25°05' a 8 = 28°40' a 9 = 30°00' (sub-chord = 2^sinl°20' = 2x400sinl°20' = 18-62ft) Example 11.10 A curve of radius 1000 ft is to be set out connecting 2 straights as shown in Fig. 11.28. Point X is inaccessible and BC is set out and the data shown obtained. Assuming the chainage at B is 46 + 47*3 ft, calculate sufficient data to set out the curve by deflection angles from the tan- gent by chords of 50 ft based on through chainage. CIRCULAR CURVES 583 0902-69} 7i / 950) T, (599709) Fig. 11.28 (derived values in brackets) In triangle BXC, (Fig. 11.28) BX = 1050-4 sin 54°30' cosec60° = 987-44 ft XC = 1050-4 sin 65°30' cosec60° = 1103-69ft In triangle OT^X, Tangent length T,X = 1000 tan 60° = 1732-05 ft .*. BT, = T,X- BX = 1732-05 - 987-44 = 744-61 ft Similarly CT 2 = 1732-05 - 1103-69 = 628-36 ft Chainage r, = 4647-30 - 744-61 = 3902-69 ft Length of curve = R <£ = 1000 x 2-094 = 2094-40 ft Chainage T z = 3902-69 + 2094-40 = 5997-09 ft 584 SURVEYING PROBLEMS AND SOLUTIONS Length of chords, C, = 3950-3902-69 = 47-31 ft C 2 = C n _, = 50-0 ft C n = 5997-09-5950 = 47-09 ft Deflection angles (Eq. 11.40) 206 265 x 50 for C, = 50 ft, a 2 = — - — = 5156-62" = 1°25'57 for C, = 47-3 ft, a, = ■ " " = 4879' for C n = 47-1 ft, a n = 2x 1 L000 5156-6 x47-3 50 5156-6 x47-l 50 = 4856" (say 1°26'00") = 1°21 , 19" (say 1°21'20") = 1°20 , 56" (say 1°21'00") Check 4879 + 40 x 5156-62 + 4856 = 216 000" = 60°00 , 00" Example 11.11 Assuming that Example 11.21 is to be set out using angular values only, calculate the deflection angle from T, and 7 2 suitable for use with a 20" theodolite, Fig. 11.29. Angles at T may be calculated as follows from the previous calculations. Fig. 11.29 a, = 1°21'19" + <x 2 = 1°25'57" + a 3 = 1°25'57" + a 4 = 1°25'57" + a s = 1°25'57" Accumulative deflection angle say 1°21'20" 2°47'20" 4°13'20" 5°39'20" 7°05'00" etc. CIRCULAR CURVES 585 Angle at T 2 , by Eq. (11.44): j8, = 360 -60 + a, = 301°21'20" B 2 = 300 +a,+ a 2 = 302°47'20" B 3 = 300 + a, + a 2 + a 3 = 304°13'20" etc. Example 11.12 Fig. 11.30 shows the centre lines of existing and proposed roadways in an underground shaft siding. It is proposed to connect roadways A-B and no. 1 shaft - C respectively by curves BD and CD, each 100 ft radius, and to drive a haulage road from D in the direction DE on a line tangential to both curves. B and C are tangent points of the respective curves and D is a tangent point com- mon to both curves. N!1 shaft SKetch-not to scale Fig. 11.30 Co-ordinates of A N 2311-16 ft, E 2745-98 ft. Co-ordinates of no. 1 shaft N 2710-47 ft, E 3052-71 ft. Bearing of no. 1 shaft - C 186°30'00". Distance of no. 1 shaft — C 355 ft. Bearing of roadway A — B 87°23 50 . Calculate the distance from A to the tangent point B of the curve BD, the co-ordinates of B and the bearing of the proposed roadway DE. (M.Q.B./S) 586 SURVEYING PROBLEMS AND SOLUTIONS > N! 1 shaft nB7* 23' 50" §_ Bearing 177°23'50" 100ft Bearing 096 # 30' 100ft Fig. 11.31 Construction (Fig. 11.31) Draw AX 100 ft perpendicular to AB Join X0 2 ; 0,0 2 . Co-ordinates of C AE^SSSsine^O' N 11 no. i AN,_ c 355cos6°30' 3052-71 -40-19 3012-52 2710-47 -352-72 2357-75 Co-ordinates of 0, E c 3012-52 AE^lOO sin85°30' +99-36 E , 3111-88 N c 2357-75 AN ^100 cos 85°30' - 11-32 N a 2346-43 CIRCULAR CURVES 587 Co-ordinates of X E A 2745-98 AE4^100sin2 o 36'10" + 4*54 E x 2750-52 N^ 2311-16 N^ 100 cos 2°36 ' 10" -99-90 Nv 2211-26 _. 3111-88 - 2750-52 _, 361-36 Bearing XO. = tan MAt . Af% — tztt~Z7 = tan T^rTZ & ' 2346-43 - 2211-26 135-17 = N 69°29'29" E Length XO, = 361-36 cosec69°29'29" = 385-81 ft In triangle X0,0 2 , XO, sin 0, XO. = sinO 2 'r»/\» Bearing X0 2 = 87°23'50' XO t = 69°29'29" /. Angle X = 17°54'21" 0, X sin X 385-81 sin 17°54'2l" sin0 2 - — jg— - — 2 = (36°22'33") or 180 - 36°22'33" = 143°38'27" 385-81 sin (180 - 143°38'27" - 17°54'21") Thus XO, = 2 sinl43°38'27" 385-81 sin 18°27'12" sin36°22'33' Co-ordinates of B = 205-91 ft = AB E A 2745-98 A E AB 205-91 sin 87°23'50" +205-70 E B 2951-68 N4 2311-16 Mi AB 205-91 cos87°23'50" + 9-35 N fl 2320-51 588 SURVEYING PROBLEMS AND SOLUTIONS Bearing X0 2 = AB = 087°23'50" Angle X0 2 0, = 143°38'27" 231°02'17" - 180° Bearing 2 0, 051°02'17" + 90° Bearing DE 141°02'17" Ans. AB 205-91 ft. Co-ordinates of B, E 2951-68 N 2320-51. Bearing of DE, 141°02'17". Exercises 11(c) 12. AB and CD are straight portions of two converging railways which are to be connected by a curve of 1500 ft radius. The point of intersection of AB and CD produced is inaccessible. X and Y are points on AB and CD respectively which are not intervisible and the notes of a theodolite traverse from one to the other are as follows: Line Horizontal Angle Distance (ft) Xa AXa 260° 10' 160 ab Xab 169°00' 240 be abc 210°30' 300 cY bcY 80°00' 180 YC cYC 268°40' _ Calculate the apex angle and the position of the start and finish of the curve relative to X and Y . (M.Q.B./S Ans. 91°40'; T,X 1234-42 ft; T 2 Y 780-09 ft) 13. To locate the exact position of the tangent point T 2 of an exist- ing 500 ft radius circular curve in a built-up area, points a and d were selected on the straights close to the estimated positions of the two tangent points T, and T 2 respectively, and a traverse abed was run between them. Station Length (ft) Deflection Angle a 9°54' R b ab 178 19°36' R c be 231 30°12' R d cd 203 5° 18' R The angles at a and d were relative to the straights. Find the distance T 2 d. (L.U. Ans. 34- lft) 2° 36-6' 72 - 4° 06-6' 7° 06-6' 75 - 8° 36-6' l°36-6' T.P. 13° 00') CIRCULAR CURVES 589 14. Two straights AI and BI meet at / on the far side of a river. On the near side of the river, a point E was selected on the straight AI, and a point F on the straight BI, and the distance from £ to F measured and found to be 3*40 chains. The angle, AEF, was found to be 165°36' and the angle BFE 168°44' . If the radius of a circular curve joining the straights is 20 chains, calculate the distance along the straights from E and F to the tangent points. (I.C.E. Ans. 3-005 ch; 2-604 ch) 15. The centre line of a proposed railway consists of two straights joined by a 3° curve. The angle of deflection between the straights is 26° and the chainage (increasing from left to right) of their intersection is 7367 ft. Calculate the deflection angles from the tangent for setting out the circular curve from the first tangent point by pegs at every 100 ft chain- age and check on to the second tangent point. (I.C.E. Ans. peg 70- l°06-6' 71- 73 - 5° 36-6' 74 - 76 - 10° 06-6' 77 - 16. Outline three different methods for setting out a circular curve of several hundred feet radius using a chain and tape and without using a theodolite. Sketch a diagram for each method and quote any formulae used in the calculations associated with each method. The centre line of a certain length of a proposed road consists of two straights with a deflection angle of 30°00' and joined by a circular curve of 1000 ft radius; the chainage of the tangent point on the first straight is 3630 ft. The curve is to set out by deflection angles from this tangent point using a theodolite which reads to 20", and pegs are required at every 100 ft of through chainage and at the second tangent point. Calculate these deflection angles and any other data that could be used in the field for checking the position of the second tangent point. (I.C.E. Ans. 2°00'20"; 4°52'20"; 7°44'00" 10°36'00"; 13°28'00"; 15°00'00") 17. The tangent length of a simple curve was 663-14 ft and the de- flection angle for a 100 ft chord 2° 18' . Calculate the radius, the total deflection angle, the length of the curve and the final deflection angle. (L.U. Ans. 1245-5 ft; 56°03'50"; 1218-7 ft; 28°01'55") 18. (a) What is meant by the term 'degree of curve' (D)? State the advantages, if any, of defining a circular curve in this way. Show how the degree of curve is related to the radius of the curve (R). 590 SURVEYING PROBLEMS AND SOLUTIONS (b) Two straights EF and FG of a proposed road intersect at point F. The bearings of the straights are: EF 76°12' FG 139°26' The chainage of E is 11376-0 (113 + 76-0) and the distance EF is 2837-6(28 + 37-6). Calculate the chainages of the tangent points and prepare a table of the deflection angles from the tangent point on GF for pegs at whole 100 ft chainages for a 5° curve connecting the two straights. Explain with the aid of a diagram how you would set out this curve. (R.I.C.S./L Ans. R = 1146 ft; 10670-3; 11935-4) 19. A circular curve XT of 1000 ft radius joins two straights AB and BC which have bearings of 195° 10' and 225°40' respectively. At what chainage from X measured along the curve, will the curve be nearest to point B? If this point of nearest approach to B be point W what is the bearing of WB? (R.I.C.S./G Ans. 266-2 ft; 120°25') 20. (a) The lines AB and BC are to be joined by a circular curve of radius 3000 ft. The point B of intersection of the lines is inaccess- ible. The following values have been measured on the ground: angle AMN = 146°05' , angle MNC = 149°12' , MN = 2761ft; and the chainage of M is 25342ft (measurement from A). Calculate the length of the curve and the chainage of the beginn- ing and end of the curve. (b) Deduce formulae for setting out intermediate points on the curve by using steel band, linen tape, optical square and ranging rods only. (R.I.C.S./L Ans. 3388-6; 25004-7; 28393-3) 21. A BC D is a plot of land, being part of a block. It is required to round off the corner by a circular curve tangential to the boundaries at B and C. What is the radius of the curve to the nearest tenth of a foot? Angle BAD = 90° AB = 100 ft Angle ADC = 52°38' AD = 140 ft (R.I.C.S./L Ans. R = 28-2 ft) 22. A railway boundary CD in the form of a circular arc is intersect- ed by a farm boundary BA in E. Calculate this point of intersection E. The radius of the curve is 500 ft and the co-ordinates in feet are: A E 7525-7, B E 7813-4 C E 8009-3 D E 7101-2 CIRCULAR CURVES 591 N 21951-7 N 20 163-3 N 21 179-6 N 21074-3 (Ans. E 7610-5 N 21424-7) 23. A 750 ft length of straight connects two circular curves which both deflect right. The first is of radius 1000 ft, the second is of radius 800 ft and deflection angle 27° 35' . The combined curve is to be replaced by a single circular curve between the same tangent points. Find the radius of this curve and the deflection angle of the first curve. (L.U. Ans. R = 1666-5 ft; 31°33') 24. In a level seam two roadways AB and DC are connected by roadways AD and BC. Point B is 820 ft due East of A, D is 122 ft due North of A, and C is 264 ft due North of B. It is proposed to drive a circular curve connecting BA and DC and tangential to BA, AD and DC. Calculate the radius of the curve and the distances from A and D to the tangent point of the curve on the lines AB and DC respectively. (M.Q.B./S Ans. 66'24 ft; 66'24 ft; 55'76 ft) 25. The co-ordinates of two points A and B are: E N A 0-0 399-60 B 998-40 201-40 A straight line AC bears 110°30' and intersects at C a straight line BC bearing 275° 50'. The chainage of A is 2671-62 ft. Calculate the lengths of AC and CB. The two straights are to be joined by a curve of 500 ft radius. Cal- culate the chainage of the tangent points and of B. (L.U. Ans. 377-97ft; 647'71ft; 2985'24ft; 3113'23ft; 3696'59ft) 11.8 Compound Curves Compound curves consist of two or more consecutive circular arcs of different radii, having their centres on the same side of the curve. There are seven components in a compound curve made up of two arcs: Two radii, /?, and R. Two tangent lengths, t x and t 2 An angle subtended by each arc, a 1 and a 2 . 592 SURVEYING PROBLEMS AND SOLUTIONS A deviation angle at the intersection point /, = a,+ a. At least 4 values must be known. Fig. 11.32 In Fig. 11.32, AB sin a ^' tan \ a ' + R * tan 1 a ^ sin a 2 A/ = sin <f> TJ = T,A + AI = f, sin<£ * , = /? , tan -= a , + (#, tan 2 a, + i? 2 tan-2<x 2 ) sin a. sin <fi t , sin <f> = R y tan -1 a, sin <£ + tf, tan -^ a, sin a 2 + R 2 tan -1 a 2 sin a 2 sin 2 o^ 2 sin ^ a 2 cos ^ a 2 = R , tan -_- a, (sin <f> + sin a 2 ) + /?, cos - a 2 = R, sini(<£- a 2 )2sin^(0+ a 2 ) cos -!(<£- a 2 ) cos 4(0- a 2 ) + 2&, sin 2 \ a = /?,( c °s a 2 - cos 0) + /? 2 (l-cosa 2 ) = fl,{(l - COS<£)-(l - COSOj) } + /? 2 (1 - COSCL,) = (R 2 - R, )(1 - cos o^) + R, (1 - cos 0) ^sincft = (/? 2 - /?,) versinea 2 + /^versineqS (11.46) Similarly, it may be shown that f 2 sin<£ = (/?,- R z ) versinea, + /? 2 versine<£ (11.47) CIRCULAR CURVES 593 An alternative solution is shown involving more construction but less trigonometry, Fig. 11.33. Construction Fig. 11.33 Produce arc T,T 3 of radius R, to 8 Draw 0,3 parallel to 2 T 2 BC parallel to IT 2 BD parallel to T 3 2 T, A perpendicular to 0, B T, E perpendicular to T 2 1 (produced) N.B. (a) T 3 BT 2 is a straight line. (b) T 3 0,0 2 is a straight line. (c) BD = 0,0 2 = DT 2 = R 2 -R,. T,E = AB + CT 2 = (0,8-0,4) + (DT 2 -DC) i.e. *, sin<£ = /?,-£, cos(a, + a 2 ) + \(R 2 - #i) - (K 2 - #i) cos a 2 \ = #,(1 - cos 96) + (R 2 - #i)(l - cosa 2 ) f, sin<£ = R, versinec6 + (R z -R x ) versine a 2 (11.48) By similar construction, Fig. 11.34, FT 2 = GH - HJ = (0 2 H - 2 G) - (HK - JK) t 2 sincf> = R 2 vers (f> + (/?,_ R 2 ) versa, (11.49) 594 SURVEYING PROBLEMS AND SOLUTIONS Fig. 11.34 Example 11.13. Given K, = 20 m, R 2 = 40 m, T,/ = 20-5 m, <f> = 80° 30', it is required to find length T 2 l. From Eq. (11.48), T,/ sin<£ - /?, versine^ versine a, = R 2 - R t 20-5 sin 80° 30' - 20 vers 80° 30' 40 - 20 3-520 20 then From Eq. (11.49), a 2 = 34° 31' a, = 45° 59' i.e. <f> - a z T , _ R 2 vers<ft - (R 2 - R } ) vers a i sin <p 40 vers80°30' - 20 vers 45° 59' sin 80° 30' 27-67 m Alternative solution from first principles (Fig. 11.35) Construction Join 0,/ Draw 0,P parallel to 2 T 2 0,Q parallel to IT 2 CIRCULAR CURVES 595 In triangle TJO,, 6 = tan-'^A * -t 20 — = tan ' 7,/ 20-5 Fig. 11.35 = 44° 18' j8 = 180 - (0 + <f>) = 55° 12' In triangle IPO, , IP = 0,1 cos j8 = R, cosec cos/3 = 20 cosec 44° 18' cos 55° 12' = 16-35 0,P = 0,1 sinj8 = 20 cosec 44° 18' sin 55° 12' = 23-52 2 Q = 2 T 2 - QT 2 = R 2 - 0,P = 40 - 23-52 = 16-48 In triangle 0,0 2 Q, a 2 =cos-M = cos-1^48 = 3^ 2 0,0, 20 ••• a, = 45° 59' 0,Q = 0,0 2 sina 2 = 20sin34°3l' = 11-32 m IT 2 = IP + PT 2 = IP + 0,Q = 16-35 + 11-32, = 27-67 m Example 11.14. AB and DC are the centre lines of two straight portions of a railway which are to be connected by means of a com- pound curve BEC, BE is one circular curve and EC the other. The radius of the circular curve BE is 400 ft. Given the co-ordinates in ft : B N400 E200, C N593 E 536, and the directions of AB and DC NE25°30' and NW76°30' respectively, Calculate (a) the co-ordinates of E, (b) the radius of the circular curve EC. (R.I.C.S./M ) Formulae are not very suitable and the method below shows an alternative from first principles. Bearing BC = tan"' 536 ~ 200 = 60°07' 593 - 400 BC = 193 sec 60° 07' <f> = 180 - 76° 30' - 25° 30' = 78° 00' 596 SURVEYING PROBLEMS AND SOLUTIONS E 53§i_— — ---s»*^ Fig. 11.36 In triangle S/,C, B = 60° 07' - 25° 30' = 34° 37' C = 78° 00' - 34° 37' = 43° 23' 8/, = BC sine cosecc6 = 193 sec 60° 07' sin 43° 23' cosec 78° = 272-0 ft C/, = BC sin B cosec ^ = 225-0 ft Bh = 400 tan Z§ 2 2 / 2 G = 323-9 ft /,/ 2 = 323-9 - 272 = 51-9 ft I,J = 51-9 cos 78° = 10-8 ft IgJ = 51-9 sin 78° = 50-8 ft FG = CH = 2 L = / 2 G + /,./- /,C = 323-9 + 10-8 - 225-0 = 109-7 In triangle CGH, a 2 2 22. = tan- 1 GH/CH a 2 = 2 tan"' I 2 J/CH = 2 tan"' = 2 x 24° 50' = 49° 40' a, = 78°- 49° 40' = 28° 20' 50-8 109-7 CIRCULAR CURVES 597 n triangle 0^0 2 L, 0^0 2 = 2 L coseca 2 = 109-7 cosec49°40' = 143-9 .*. R 2 = 400 - 143-9 = 256-1 to find the co-ordinates of E, Bearing BE = 25° 30' + 14° 10' = 39° 40' BE = 2 x 400 sin 14° 10' Partial Lat. = BE cos 39° 40' = 150-71 Total Lat. = N 550-7 Partial Dep. = BE sin 39° 40' = 124-98 Total Dep. = E 325-0 Iheck From Eq. (11.48), /, sin<£ = 272 sin 78° = 266-06 ft R, versine^ = 400(1 - cos 75°) = 316-83 ft (R 2 - R,) versinea 2 = (256-1 - 400) (1 - cos49°40') = -50-76 ft .*. /?, versc6 + (R 2 - /?,) vers a 2 = 316-83 - 50-76 = 266-07 ft Example 11.15. Undernoted are the co-ordinates in ft of points on the ■espective centre lines of two railway tracks ABC and DE Co-ordinates . A B E 525-32 N 52-82 C E 827-75 N 247-29 D E 10-89 S 108-28 E E 733-23 S 35-65 The lines AB and DE are straight and B and C are tangent >oints joined by circular curve. It is proposed to connect the two racks at C by a circular curve starting at X on the line DE, C being t tangent point common to both curves. Calculate the radius in chains >f each curve, the distance DX and the co-ordinates of X. (M.Q.B./S) In Fig. 11.37, Bearing AB = tan" 1 + 525-32/+ 52*82 = N 84° 15' 30" E 598 Bearing DE = tan SURVEYING PROBLEMS AND SOLUTIONS _, 733-23 - 10-89 -35-65+ 108-28 = N 84° 15' 30" E As bearings are alike, C, B and X must lie on the same straight line. Bearing BC = bearing XC = tan-1 827 ' 75 - 525-32 247-29 - 52-82 = N 57° 15' 29" E Length BC = AN sec bearing = (247-29 -52-82) sec 57°15'29" = 359-56 Angle (a) subtended at centre = 2 x angle BXE = 2 x (84° 15' 30" - 57° 15' 29") Fig. 11.37 = 54°00'02' Radius 2 C = ±BC cosecia = I x 359-56 cosec27°00'01" 2 396 ft = 6 chains Bearing DB = tan-' 525 ' 32 10-89 = N 72° 36' 41" E 52-82 + 108-28 Length DB = 161-10 sec 72° 36' 41" = 539-06 Bearing DE = 084° 15' 30" DB = 072° 36' 41" .-. Angle BOX = 11°38'49" BC = 057° 15' 29" /. Angle DBX = 15° 21' 12" In triangle DBX, DX = BX = BD sing _ 539-06 sin 15° 21' 12" sinX ~ ■ sin 27° 00' 01" BD sin D 539-06 sin 11° 38' 49" sinX sin 27° 00 '01" Corordinates of X Line DX N 84° 15' 30" 314-38 ft .'. AE = 314-38 sin 84° 15' 30" = +312-80 E D = 10-89 E^ = 323-69 AN = 314-38 cos 84° 15' 30" = + 31-45 Nn = -108-28 = 314-38 ft = 239-71 ft CIRCULAR CURVES 599 N x = -76-83 CX = BX + BC = 239-71 + 359-56 = 599-27 ft Radius 0,C = |CXcosec27°00'0l" = 1 x 599-27 cosec27°00'0l" = 659-986 ft a* 660-0 ft = 10 chains Ans. O z C = 6 chains radius O^C = 10 chains radius DX = 314-38 ft Co-ordinates of X E 323-69, S 76-83 Exercises 11(d) (Compound curves) 26. A main haulage road AD bearing due north and a branch road DB bearing N87°E are to be connected by a compound curve formed by two circular curves of different radii in immediate succession. The first curve of 200 ft radius starts from a tangent point A, 160 ft due South of D and is succeeded by a curve of 100 ft radius which termin- ates at a tangent point on the branch road DB. Draw a plan of the roadways and the connecting curve to the scale 1/500 and show clearly all construction lines. Thereafter calculate the distance along the branch road from D to the tangent point of the second curve and the distance from A along the line of the first curve to the tangent point common to both curves. (M.Q.B./S Ans. 120-2 ft; 145-4 ft) 27. The intersection point / between two railway straights T'XI and IYT is inaccessible. Accordingly, two arbitrary points X and Y are selected in the straights and the following information is obtained: Line Whole circle bearing T'XI 49° 25' IYT" 108° 40' Length XY = 1684-0 ft XY 76° 31' ChainageofX = 8562-3 ft The straights are to be connected by a compound circular curve such that the arc T'C, of radius 3000 ft, is equal in length to the arc CT", of radius 2000 ft, C being the point of compound curvature. Make the necessary calculations to enable the points T', C and T" to be pegged initially, and show this information on a carefully dimensional sketch. 600 SURVEYING PROBLEMS AND SOLUTIONS Determine also the through chainages of these points. (L.U. T'l = 1489-1 ft chainage T 8115*9 ft T'l = 1235-5 ft chainage T" 10 597-7 ft C 9356-8 ft) 11.9 Reverse Curves There are four cases to consider: (1) Tangents parallel (a) Radii equal (b) Radii unequal. (2) Tangents not parallel (a) Radii equal (b) Radii unequal. Tangents parallel (Cross-overs) (Fig. 11.38) T, T 2 , 0, 2 and /, / 2 are all straight lines intersecting at E, the common tangent point, Fig. 11.38 In Fig. 11.38, Angle BT^E = T, £/, = ^ 0, T EI = —T O E = \bIJ 2 = a = icy, = a Angle CT 2 £ Triangle T,/,fi is similar to triangle ET 2 / 2 Triangle 1,^0, is similar to triangle E0 2 T 2 Tangents not parallel (Fig. 11.39) N.B. T, T 2 does not cut 0^0 2 at E. and Fig. 11.39 CIRCULAR CURVES 601 N.B. These solutions are intended only as a guide to possible methods of approach. Tangents parallel (Fig. 11.40) Bisect T,£ and T 2 E. Draw perpendiculars PO^ and QQ 2 . Fig. 11.40 *i \-» / ! NX j — * — ^ T,E = 2T,P = 2K, sina T 2 E = 2T 2 Q = 2K 2 sina = T,T 2 - T,£ 2R Z sin a = I, T 2 - 2K, sin a XT, If R, = R z , R, = R = 2 sina T T 4 sina - *, (11.50) (11.51) Example 11.16 Two parallel railway lines are to be connected by a reverse curve, each section having the same radius. If the centre lines are 30 ft apart and the distance between the tangent points is 120 ft, what will be the radius of cross-over? In Fig. 11.41, T,A = 30 ft T X T = 120 ft In triangle T x T Z A, T.A sina = —— - = i, i z 30 120 In triangle T x PO x , R = T, 0, = T, P T, T 2 Fig. 11.41 120 sina 4 sina 4x30/120 = 120 ft Tangents not parallel, radii equal (Fig. 11.42) Construction Draw 0, S parallel to T, T 2 , PO x perpendicular to T, T 2 , Q per- pendicular to T, T 2 602 SURVEYING PROBLEMS AND SOLUTIONS In triangle T, PO x , I, P = R sin a, PO, = R cos a, = QS In triangle 2 T 2 Q, T 2 Q = R sina 2 2 Q = R cosa 2 2 S = 2 Q + QS = R cosa 2 + R cos a, = i?(cosa, + cosa 2 ) 2 S P ' sin 0^ 2 Fig. 11.42 = sin _ l /?(cosa, + cosa 2 ) 2R 0,S = 2R cos jS = T,T 2 -(T,P + QT 2 ) = T y T 2 - K(sina, + sina 2 ) T T 2 cos|3 + sina, + sina 2 (11.52) Example 11.17 Two underground roadways AB and CD are to be connected by a reverse curve of common radii with tangent points at B and C. If the bearings of the roadways are AB S 83° 15' E and CD CD S 74° 30' E and the co-ordinates of B E 1125*66 ft N 1491-28 ft, C E 2401-37 ft N 650*84 ft, calculate the radius of the curve. (M.Q.B./S) Fig. 11.43 The bearings of the tangents are different and thus the line BC does not intersect with 0»0 2 at the common tangent point although at this scale the plotting might suggest this. In Fig. 11.43, CIRCULAR CURVES 603 n . Qn , -.2401-37-1125-66 Bearing BC = tan ' —T^z—rz — « AM nn = tan" 650-84-1491-28 , 1275-71 -840-44 = S 56° 37' 23" E Length BC = 840-44 sec 56° 37' 23" = 1527-67 ft Bearing BC S 56° 37' 23" E AB S 83° 15' 00" E .-. a, = 26° 37' 37" CD S 74° 30' 00" E ctj, = 17° 52' 37" /8 = sin" 1 I (cos 26° 37' 37" + cos 17° 52' 37") = 67° 20' 39" 0,S = 2R cos 67° 20' 39" BP = R sin a, = R sin 26° 37' 37" QC = R sina 2 = R sin 17° 52' 37" BC = BP + 0,S + QC .-. 1527-67 = R sin 26° 37' 37" + 2R cos 67° 20' 39" + R sin 17° 52' 37" By Eq. (11. 50), R 1527-67 ~ sin 26° 37' 37" + 2 cos 67° 20' 39" + sin 17° 52' 37" = 1001-31 ft Tangents not parallel, radii not equal (Fig. 11.44) Fig. 11.44 Construction Join 0, T 2 . Draw 0, P perpendicular to T, T 2 T, P = /?, sin a, O^P = #, cos a, 604 SURVEYING PROBLEMS AND SOLUTIONS PT 2 = T y T 2 - T,P = T,T 2 - /?, sina, = tan '-^=- = tan (11.53) Pi 2 7, T 2 - /<! sin a, n t °' P _ R i cosa i sintf ~ sin0 In triangle 0, 2 I 2 , 0,01 = OJl + 0,T 2 2 - 20 2 T 2 0J 2 cos 0,f 2 2 R* cos 2 <x, (fl, + R 2 y = K 2 2 + sin 2 2/? 2 K, cos a, cos {90 - (a 2 - 0)1 sin0 2 22 #, 2 cos 2 a,- 2#,i? 2 cosa, sin(a 2 -0)sin0 /v. + 2R.R? + Rf = /v., H ; — 2 ~: sin 6 R,(R, cos 2 a, - 2R 2 cos a, sin(a 2 - 0) sin 0) /?,(*, + 2K 2 ) = ~i^ 1 2 -—rr ~ sin u 2R 2 sin {sin + cos a, sin(a 2 - 0)} = R,(cos 2 a, -sin 2 0) R = /? 1 (cos 2 a 1 - sin 2 0) 2 2 sin {sin + cosa,sin(a> - 0)} Example 11.18 Two straights AB and CD ate to be joined by a cir- cular reverse curve with an initial radius of 200 ft, commencing at B. From the co-ordinates given below, calculate the radius of the second curve which joins the first and terminates at C. E N Co-ordinates (ft) A 103-61 204-82 B 248-86 422-62 C 866-34 406-61 D 801-63 141-88 . (R.I.C.S./M) A 103-61 El 204-82 N Fig. 11.45 CIRCULAR CURVES 605 In Fig. 11.45, Bearing AB = tan- g^*|^j|^ = N 3 3°42'00" E = 033° 42*00' i r\r\H DC = tan-' ^'^-^l'S - N 13°44'00" E = 013°44'00 (406*61 -141-88) BC = tan"' (866 ' 34 ~ 248 ' 86) = S 88°30'50" E = 091° 29' 10" (406-61-422-62) Length BC = 617-48/cos88°30'50" = 617-69 ft (T, T 2 ) Angle a, = 91° 29' 10" - 033° 42' 00" = 57° 47' 10" a = 91 o 29'10"-013°44'00" =77° 45' 10" 2 By Eq. (11.53), R,cosa, , 200 cos 57° 47' 10" v = tan '-srs s ^— = tan „«„-.,„ nnn ... fw0 A n> m» TJ 2 - R, sina, " 617-48 - 200 sin 57° 47' 10 = 13° 22' 20" By Eq. (11.54), 200 (cos 2 57° 47' 10" - sin 2 13° 22' 20") 2 ~ 2 sin 13° 22' 20" {sin 13° 22' 20" + cos 57° 47' 10" sin(77°45'10" - 13° 22' 20 = 140-0 ft Exercises 11(e) (Reverse curves) 28. A reverse curve is to start at a point A and end at C with a change of curvature at B. The chord lengths AB and BC are respec- tively 661-54 ft and 725*76 ft and the radius likewise 1200 and 1500 ft. Due to irregular ground the curves are to be set out using two theodolites and no tape and chain. Calculate the data for setting out and describe the procedure in the field. (L.U. Ans. Total deflection angles, 16°; 14°; Setting out by 1° 11' 37" deflections) 29. Two roadways AB and CD are to be connected by a reverse curve of common radius, commencing at B and C. The co-ordinates of the stations are as follows: A 21642-87 m E 37 160*36 m N B 21 672*84 m E 37 241*62 m N C 21 951*63 mE 37350*44 m N If the bearing of the roadway CD is N 20° 14' 41" E, calculate the radius of the curve. (Ans. 100*0 m) 606 SURVEYING PROBLEMS AND SOLUTIONS 30. CD is a straight connecting two curves AC and DB. The curve AC touches the lines AM and CD; the curve DB touches the lines CD and BN. Given: Bearing MA = 165° 13' BN = 135° 20' Radius of curve AC = 750 m DB = 1200 m Co-ordinates A E + 1262*5 m N - 1200-0 m B0 Calculate the co-ordinates of C and D. (Ans. C E + 1562-54 m N - 1358*58 m; D E + 57-32 m N - 53-09 m) 31. Two parallel lines which are 780 m apart are to be joined by a reverse curve ABC which deflects to the right by an angle of 20° from the first straight. If the radius of the first arc AB is 1400 m and the chainage of A is 2340 m, calculate the radius of the second arc and the chainages of B and C. (Ans. 1934 m; 2828 m; 3503*8 m) 32. Two straight railway tracks 300 ft apart between centre lines and bearing N 12° E are to be connected by a reverse or 'S* curve, starting from the tangent point A on the centre line of the westerly track and turning in a north-easterly direction to join the easterly track at the tangent point C. The first curve AB has a radius of 400 ft and the second BC has a radius of 270 ft. The tangent point common to both curves is at B. Calculate (a) the co-ordinates of B and C relative to the zero origin at A (b) the lengths of the curves AB and BC. (M.Q.B./S Ans. (a) B E 244-53 N 288*99 ft C E 409*58 N 484*05 ft (b) 394*30 ft; 266*15 ft) Bibliography THOMAS, W.N., Surveying (Edward Arnold) CLARK, D., Plane and Geodetic Surveying (Constable) DAVIES R.E., FOOTE F.S., and KELLY, j.w., Surveying Theory and Practice (McGraw-Hill) SANDOVER, J. A., Plane Surveying (Edward Arnold) BANNISTER, A., and RAYMOND, S., Surveying (Pitman) JAMESON, A.H., Advanced Surveying (Pitman) SALMON, V.G., Practical Surveying and Fieldwork (Griffin) JENKINS, R.B.M., Curve Surveying (Macmillan) KISSAM, P., Surveying for Civil Engineers (McGraw-Hill) SEARLES, W.H. and IVES, H.C., rev. by P. KISSAM, Field Engineering, 22nd Ed. (John Wiley) HIGGINS, A.L., Higher Surveying (Macmillan) 12 VERTICAL AND TRANSITION CURVES 12.1 Vertical Curves Where it is required to smooth out a change of gradient some form of parabolic curve is used. There are two general forms, (a) convex or summit curves, (b) concave, i.e. sag or valley curves. The properties required are: (a) Good riding qualities, i.e. a constant change of gradient and a uniform rate of increase of centrifugal force. (b) Adequate sighting over summits or in underpasses. The simple parabola is normally used because of its simplicity and constant change of gradient, but recently the cubic parabola has come into use, particularly for valley intersections. It has the advant- age of a uniform rate of increase of centrifugal force and less filling is required. Gradients are generally expressed as 1 in x, i.e. 1 vertical to x horizontal. For vertical curve calculations % gradients are used : e.g. linx = m% x 1 in 5 = 20% Gradients rising left to right are considered +ve. Gradients falling left to right are considered - ve. The grade angle is usually considered to be the deflection angle = difference in % grade. By the convention used later, the value is given as q%-p%, Fig. 12.1. -Q -p = -(q+p)v. q-(-p)**(q+p)»l. 608 SURVEYING PROBLEMS AND SOLUTIONS 12.2 Properties of the Simple Parabola y = ax 2 + bx + c & = lax + b dx 2a for max or mm (12.1) (12.2) dJz , = 2a, i.e. constantrateofchangeofgradient (12.3) If a is +ve, a valley curve is produced. If a is -ve, a summit curve is produced. The value of b determines the maximum or minimum position along the x axis. The value of c determines where the curve cuts the y axis. The difference in elevation between a vertical curve and a tangent to it is equal to half the rate of change of the gradient x the square of the horizontal distance from the point of tangency, Fig. 12.2. Fig. 12.2 If y = ax 2 + bx + c, -¥- = 2 ax + b (grade of tangent) dx When x = 0, grade of tangent = +b value of y = + c In Fig. 12.2, if the grade of the tangent at x = is b, VERTICAL AND TRANSITION CURVES 609 QR = b x x i.e. + bx RS = +c PQ = ax 2 as y = PQ + QR + RS = ax 2 + bx + c The horizontal lengths of any two tangents from a point to a verti- cal curve are equal, Fig. 12.3. PB = ax? = ax\ (1) . . x x — x z i.e. the point of intersection B of two gradients is horizontally mid- way between the tangent points A and C. A chord to a vertical curve has a rate of grade equal to the tangent at a point horizontally midway between the points of intercept, Fig. 12.3. From (1), AA X = CC X = ax 2 AA, is parallel to CC, ACC^A % is a parallelogram AC is parallel to i4,C,. Fig. 12.3 12.3 Properties of the Vertical Curve From the equation of the curve, y = ax 2 + bx + c as seen previ- ously. C A P^ c, *i *i B *a Fig. 12.4 610 SURVEYING PROBLEMS AND SOLUTIONS In Fig. 12.4, AB = ax 2 BC = bx (N.B. For summit curves, a is negative) .'. the level on the curve at A = ax 2 + bx DT 2 = al z = ET Z - ED = JL - JpL = Kg-p) 200 200 200 Half the rate of change of gradient (N.B. a will be negative when p > q) a = g-P 200/ Length of curve (12.4) ' = 4^ ( 12 - 5 ) 200 a It is common practice to express the rate of change of gradient 2a as a % per 100 ft. .'. the horizontal length of curve may be expressed as 100 x grade angle / % rate of change of grade per 100 ft 100 (q - p) 2a%per 100 Distance from the intersection point to the curve IF = KT, = FG = JT 2 2 (12.6) ■ •©" L. x q ~ p = l ( q ~ p ) (12.7) 4 200/ 800 Maximum or minimum height on the curve H = ax 2 + bx = (q-p)x 2 + p±_ 200 / 100 for max or min dH = (q - p) x | p = Q dx 100/ 100 then x = ~ p/ = pl (12.8) <?-P P- Q VERTICAL AND TRANSITION CURVES 611 Example 12. 1 A parabolic vertical curve of length 300 ft is formed at a summit between grades of 0*7 per cent up and 0*8 per cent down. The length of the curve is to be increased to 400 ft, retaining as much as possible of the original curve and adjusting the gradients on both sides to be equal. Determine the gradient. (L.U.) m. iwiw <-'v£. y.».*..-ry, a _ q - p 200/ -0-8 - - 0-7 1-5 " 200 x 300 60000 If the gradients are to be made e qual, p = q. P + Q = 200 a/ 2p = 200 x 400 1-5 60000 P = 1% 12.4 Sight Distances (s) 12.41 Sight distances for summits (1) s > / In Fig. 12.5, Let / = horizontal length of vertical curve sight distance AC = height of eye above road at A OL = height of object above road at distance of vehicle from tangent point T, distance of object from tangent point T 2 TJ = JP- + 12- = -L (n + q) 2 200 200 200 KP qJ I s A, <*, (1)s>< Fig. 12.5 Principles of sight distances 612 AB Similarly, SURVEYING PROBLEMS AND SOLUTIONS . T 2 J p + q tana = ■==-— = nnn T,J 200 *£ d,tana = d 'P *<? + «> _ d ' (2p p- 100 ' 100 200 200 -9) = J±(p-q) 200 MO - d tan a - hi - dz ^P + g) £^ _ A. / n , . 2 /rt MU - d 2 tana - — _ 200 100 ~ 200 {p + q ~ 2q) 200 ft, = AB + BC = Jl-(p-q) + a(i) 2 200 V2/ but a is negative, i.e. a = ^ — Ei 200/ *' - 200 tP *' 200/ X 4 - A. r D - a) + /( P~^ - P ~ g T4rf + /I ~ lOO tP ^ + ^00 8O0T L ' + J Similarly, h 2 = OM + BC = J=LjLi[4d a+ Z] , _ . . 800ft, - /(p-g) 800ft 2 - /(p-g) _ _ d} + dz = __ + __ 400(6, +h 2 ) - Up-q) 2(p - q) 2s -21 = 400 ^ + *a) _ , P - q I = 2s - 4 00(^i +^2) P - q (12.9) If s = /, / = 2/ - 4 00(^i + *fc) P - 9 / = 400(ft« + ^) (1210) P - 9 (2) s < / In Fig. 12.6, ft, = ad] .-. d. = )!h h, = adt VERTICAL AND TRANSITION CURVES yjh 2 613 *. dp = y/a V" (2) S<1 But p - a a = — 200/ 200/ tV^i +\f h z]' BRSiSTO! Fig.12.6 / = p - q S 2 (p - q) N.B. If h x = h 2 = h: (i) s > I (12.11) (ii) s = / (iii) s < I I = 2s - = 2s - I = / = 400 x 2ft p - q 800 ft p - q 800/2 p - q s\p - q) 200.[2(V/i)] 2 s 2 (p - g) 800 h (12.12) (12.13) (12.14) 12.42 Sight distances for valley curves Underpasses Given: clearance height H, height of driver's eye above road /i,, height of object above road h 2 , sight length s, gradients p% and q%. (1) s > I Let the depth of the curve below the centre of the chord AD (the distance between the observer and the object) be M, Fig. 12.7. GL _ DE s tana = GL is GJ + JI + IL 4 200 By Eq. (12.4), a = _ q - P 200/ 614 SURVEYING PROBLEMS AND SOLUTIONS itiie&ft Fig. 12.7 Sight distances for valley curves and from above, DE = 2GL Then -^- + -21 = -*- (p + q) 200 200 200 GL = M + /(< ?-P> + ^P = s <P + g) 4 x 200 200 400 / = \ s(p + q) _ JW _ M ] ^ <7 - P L 400 200 J ' 800 2sp + 2sq - 4sp ~ 800 M q - P 2s(q-p) - 800JW = 2s - q - p 800 M If AL = LE, then M = H q-p h, + h 2 and / = 2s - 800 Iff - fc i+M (12.15) (2) s 4 I In Fig. 12.8, let the depth of the curve below the centre of the chord T, T 2 = M u = ml = fo ~ p) s * 4 800/ /. / = s2 ( q ~ P^ 800 M VERTICAL AND TRANSITION CURVES '<q-p) 615 (12.16) 800 (H-hl±hlj I = If S = /, q - p (2) S« (12.17) Fig. 12.8 12.43 Sight distance related to the length of the beam of a vehicle's headlamp (1) s > I In Fig. 12.9, the height of the beam = h is at A, the beam hits the road at T 2 , the angle of the beam is $° above the horizontal axis of the vehicle. S>1 HB, s, Is assumed equal to s Fig. 12.9 Sight distance related to the beam of a headlamp In triangle BT 2 0, i.e. T 2 Q = sO al z - h = s6 616 SURVEYING PROBLEMS AND SOLUTIONS But a = q ~ P 200/ (< ?-P> /2 = sd + h 200/ / = 200(s^) Q Practice suggests that 6 = 1° h = 2-5 ft Then / = 200(0-0175 s+ 2' 5) 3'5s + 500 9 - P <7 - P (2) s«Z As before, as 2 - h = sd 200/ ~ s6 + h I = s2 (g ~ P) 200(s<9 + a; If 0=1° and h = 2-5 ft (12.19) 200(s<9 1° + h) and s 2 (g- -P) (12.20) / = J ^ ~ ^ (12 21) 3-5 s + 500 K ' Example 12.2. The sag vertical curve between gradients of 3 in 100 downhill and 2 in 100 uphill is to be designed on the basis that the headlamp sight distance of a car travelling along the curve equals the minimum safe stopping distance at the maximum permitted car speed. The headlamps are 2'5ft above the road surface and their beams tilt upwards at an angle of 1° above the longitudinal axis of the car. The minimum safe stopping distance is 500 ft. Calculate the length of the curve, given that it is greater than the (L.U.) sight distance. p = -3% q = +2% s = 500 ft By Eq. (12.21), / = s 2 (q- P) 500 2 x (2+3) 500 x 5 3*5 s + 500 500 x 3*5 + 500 3-5 + 1 = 2500 4-5 555- 5 ft 12.5 Setting-out Data Gradients are generally obtained by levelling at chainage points, e.g. A,B,C and D, Fig. 12.10. VERTICAL AND TRANSITION CURVES 617 +p<y ~~^~^^qVo ^( r, '/2/ X d-x '••^^7' 2 d 1/2/ Si. A St.S St.C Levels taken at stations A,B,C and Fig. 12. 10 Setting out data Chainage and levels of I, T, and T 2 . Level of / = level of B + £— 100 = level of c + \Za Solving the equation gives the value of x. Chainage of / = chainage of B + x Chainage of T, = chainage of / - - / Chainage of T 2 = chainage of / + ^ / Level of T, = level of / - Level of T = level of / - lp 200 200 Levels on the curve (Fig. 12.11) st.o (12.22) (12.23) (12.24) (12.25) (12.26) (12.27) A/ . r~ter~~— — jp<^ r^~ i r v^i !c -<7% (a) Summit BC = tox-ax 2 Fig. 12.11 Levels on the curve (b) Valley BiC y =-bx*ax 2 618 SURVEYING PROBLEMS AND SOLUTIONS Levels on the tangent at A = level of T, + bx (12.28) where b = ±p/100. Difference in level between tangent and curve = ±ax 2 (12.29) Levels on the curve at B = level of tangent + difference in level between curve and tangent = T, ± AC + AB (12.30) Check on computation (a) Tangent level is obtained by successive addition of the dif- ference in level per station. (b) Successive curve levels should check back to the level obtain- ed by the spot level derived from y = + ax 2 ± bx (c) The final value of the check on T 2 will prove that the tangent levels have been correctly computed, though the curve levels may not necessarily be correct. In order to define the shape of the curve, the values of a and b in the formula must be in some way obtained, p and q will always be known. b = p/100 Either l,s or a must be given, and if the value of s is required, the height of the vehicle (h) must be known. For general purposes this is taken as 3*75 ft, and Ministry of Transport Memoranda on recom- mended visibility distances are periodically published. Example 12.3. As part of a dual highway reconstruction scheme, a line of levels were taken at given points on the existing surface. Reduced level Chainage A 104-63 20 + 75 B 109-13 22 + 25 C 107-29 25 + 50 D 103-79 27 + 25 If the curve, based on a simple parabola, is designed to give a rate of change of gradient of 0-6% per 100 ft, calculate: (a) the length of the curve /, (b) the chainage and level of the intersection point, (c) the chainage and level of the tangent points, (d) the level of the first three chainage points on the curve (i.e. stations 100 ft apart based on through chainage), (e) the length of the line of sight s to a similar vehicle of a driver 3 ft 9 in. above the road surface. (N.B. s < /) VERTICAL AND TRANSITION CURVES 619 104-63 2075 109-13 2225 107-29 103-79 levels 2550 2725 chainage Fig. 12. 12 (a) Gradient AB = (109-13 - 104-63) in (2225 - 2075) = 4-50 ft in 150 ft i.e. +3% CD = (107-29 - 103-79) in (1725 - 2550) = 3-50 ft in 175 ft i.e. -2% By Eq. (12.6), Length of curve = 100(3 + 2) = 333.33 ft 0-6 (b) By Eqs. (12.22/23), Level of / = 109-13 + — 100 = 107-29 + 2 ( 325 ~*) 100 Solving for x, x = 93-20 ft 325 - x = 231 -80 ft Level of / = 109-13 + 3 x 93 ' 2 = 111-93 also 100 = 107-29 + 2_*-?31± 100 = 111-93 {check) Chainage of / = Chainage of B + x = (22 + 25) + 93-20 = 23 + 18-20 (c)By Eqs. (12.24/25), Chainage of T, = " 2318-20 - 1/2 = 2318-20 - 416-67 = 1901-53 ft Chainage of T 2 = 2318-20 + 416-67 = 2734-87 ft 620 SURVEYING PROBLEMS AND SOLUTIONS By Eqs. (12.26/27), Level of T, = 111-93 - 3 x 833 ' 33 = 99-43 ft 1 200 Level of T 2 = 111-93 - 2 x * 33 ' 33 2 200 103-60 ft (d) Setting-out data ax 2 Point Chainage Length (x) Tangent Level u Formation (0-3xlO-V) Level T, 1901-5 99-43 99-43 + 2-955 2000-0 98-5 102-385 -0-291 102-09 + 3-000 2100-0 198-5 105-385 -1-181 104-20 + 3-000 2200-0 298-5 108-385 -2-675 105-71 (e) Line of sight (s) ,._o»/>i In Fig. 12.13, but Fig. 12 13 h , = adf d - IEL 1 - Va h 2 = a d z d 2 = M 2 V a s = d, + d 2 V7 + 'V a h, = h z a = 0-3 x 10" s = 2 IK = 200 A /S? 'V a ^0-3 = 707 ft Example 12.4. A 6% downgrade on a proposed road is followed by a 1% upgrade. The chainage and reduced level of the intersection point of the grades is 2010 ft and 58-62 ft respectively. A vertical parabolic curve, not less than 250 ft long, is to be designed to connect the two grades. Its actual length is to be determined by the fact that at chain- age 2180 ft, the reduced level on the curve is to be 61-61 ft to provide adequate headroom under the bridge at that point. VERTICAL AND TRANSITION CURVES 621 Calculate the required length of the curve and also the chainage and reduced level of its lowest point. (R.I.C.S.) In Fig. 12.14, Thus, Fig. 12.14 let x -I = 170ft 2 level at / = 58*62 tangent level at Y = 58*62 - (170 x 0*06) = 58-62 - 10*20 = 48-42 ft curve level at X = 61*61 ft .-. XY = 13* 19 ft ax 2 = 13*19 13* 19 i.e. a = Also, Thus al 2 = -EL + JL = JL(p +q) 200 200 200^ Hf a = P + = -J— 200/ 200/ 13-19 200/ 7x'< Thus i.e. Solving for x, x - 7x' 200 x 13*19 170 7x z 2638 5276 7x2 - 5276 x = 170 x 5276 + 5276 ± V15276 2 - 4 x 7 x 170 x 5276 j 14 = 5276 ± 1649*9 = 494 . 7 or 259 . Q ft 14 622 SURVEYING PROBLEMS AND SOLUTIONS i.e. / 2 324-7 or 89-0 ft / = 650 ft (length must not b« 250 ft) To find the minimum height on the curve, H = aX z - bX _ (p + q) X Z 200/ pX 100 dH dX _ 2(p + q) X 200/ p 100 .-. X pi _iL = p + q 557-1 ft 6 x 650 Chainage of minimum height = 2010 + 557-1 - —— = 2242-1 ft Level at minimum height: Level at 7, = 58- 62 + 0-06 x (325) = 78- 12 ft Level of tangent = 78*12 - 0-06 x (557- 1) = 44*70 i2 aX 2 = L* 55 !'i = 16-71 ft 200 x 650 Level on curve = 44-70 + 16-71 = 61-41 ft Example 12.5. On the application of the cubic parabola for valley curves. A valley curve of length 400 ft is to be introduced into a road to link a descending gradient of 1 in 30 and an ascending gradient of 1 in 25. It is composed of two cubic parabolas, symmetrical about the bi- sector of the angle of intersection of the two straights produced. The chainage of the P.I. of the straights is 265 + 87 ft, and its re- duced level 115-36 ft. Calculate: (i) The reduced levels of the beginning, the mid-point and the end of the curve. (ii) The chainage and reduced level of the lowest point of the curve. (iii) The reduced level at chainage 267 + 00 ft. VERTICAL AND TRANSITION CURVES 623 (The formula for the cubic parabola relating to the curve of total length L and terminal radius R is 3 \ y = _*_] (N.U.) 6RL/ 263+87 Fig. 12. 15 From the gradients, Fig. 12.15, 0, = cot _, 30 = 1°54'33" 0, = cot _1 25 = 2°15'27" <f> = |[180- (0, + M = 87°55'00" 6KL i£l 6RL From the equation, Y dy dx and if x = L = 400/2, then K = 100 tan<£ = 100 x 27-49 = 2749 ft cot<£ At I, y = 6R 40000 6 x 2749 - 2-43 (i) As the gradients are small, the values of y are assumed vertical. .*. Level of curve at the mid point = 115*36 + 2* 43 = 117*79 ft Level at T, = 115*36 + ^ = 115-36 + 6*67 = 122*03 ft Level at T 2 = 115*36 + ^ = 115*36 + 8-00 - 123*36 ft (ii) Level at the lowest point P (relative to T,) — x x 30 + 6RL dL = -_L + _*_ 30 2/?L dx 2RL 30 = (for min value) 2 x 2749 x 200 30 x = 191 ft from I, 624 SURVEYING PROBLEMS AND SOLUTIONS The chainage of the lowest point P = chainage of T, + x = (263 + 87) + (1 + 91) = 265 + 78 ft Level of P = - ^ + ^ - + 122-03 30 6 x 2749 x 200 = -6-37 + 2-11 + 122-03 = 117-77 ft (iii) At chainage 267 +00, i.e. 87ft from T 2 , 87 y = 2i = 0-199 ft, i.e. 0-20 ft 6 x 2749 x 200 .'. Level at 267 + 00 = 123-36 - -^ + 0-20 25 = 120-08 ft Exercises 12(a) 1. An uphill gradient of 1 in 100 meets a downhill gradient of 0*45 in 100 at a point where the chainage is 61+00 and the reduced level is 126 ft. If the rate of change of gradient is to be 0*18 % per 100 ft, prepare a table for setting out a vertical curve at intervals of 100 ft. (I.C.E. Ans. 121-97, 122-88, 123-61, 124-16, 124-53, 124-72, 124-73, 124-56, 124-19) 2. (a) A rising gradient of 1 vertically to 200 horizontally, is to be joined by a rising gradient of 1 in 400 by a 400 ft long parabolic curve. If the two gradients meet at a level of 365-00 ft A.O.D., tabulate the levels on the curve at 50 ft intervals. (b) Recalculate on the basis that the first gradient is falling and the second likewise falling in the same direction. (L.U. Ans. 364-00, 364-24, 364-47, 364-68, 364-87, 365-05, 365-22, 365-37, 365-50; 366-00, 365-76, 365-53, 365-32, 365-13, 364-95, 364-78, 364-63, 364-50) 3. A rising gradient, g,, is followed by another rising gradient g 2 (g 2 less than g,). These gradients are connected by a vertical curve having a constant rate of change of gradient. Show that at any point on the curve, the height y above the first tangent point A is given by y = g\X - ^ gl ~ ? f z where x is the horizontal distance of the point from A y and L is the horizontal distance between the tangent points. Draw up a table of heights above A for 100 ft pegs from A, when g, = +5%, g 2 = +2% and L = 1000 ft. At what horizontal distance from A is the gradient +3%? VERTICAL AND TRANSITION CURVES 625 (I.C.E. Ans. 4-85, 9-40, 13-65, 17-60, 21-25, 24-60, 27-65, 30-40, 32-85, 35-00, 667 ft) 4. A rising gradient of 1 in 100 meets a falling gradient of 1 in 150 at a level of 210*00. Allowing for headroom and working thickness, the vertical parabolic curve joining the two straights is to be at a level of 208-00 at its midpoint. Determine the length of the curve and the levels at 100 ft intervals from the first tangent point. (L.U. Ans. 960 ft; 200-40, 206-11, 206-85, 207-42, 207-81, 208-00, 208-03, 208-07, 207-94, 207-63, 207-15, 206-80) 5. On a straight portion of a new road, an upward gradient of 1 in 100 was connected to a gradient of 1 in 150 by a vertical parabolic summit curve of length 500 ft. A point P, at chainage 59 100 ft on the first gradient, was found to have a reduced level of 45-12 ft, and at point Q, at chainage 60000 ft on the second gradient, of 44*95 ft. (a) Find the chainages and reduced levels of the tangent points to the curve. (b) Tabulate the reduced levels of the points on the curve at inter- vals of 100 ft from P at its highest point. Find the minimum sighting distance to the road surface for each of the following cases: (c) the driver of a car whose eye is 4 ft above the surface of the road. (d) the driver of a lorry for whom the similar distance is 6 ft. (Take the sighting distance as the length of the tangent from the driver's eye to the road surface.) (L.U. Ans. (a) 59 200, 46-12, 59 700, 46*94 (b) 46-12, 46-95, 47-45, 47-62 (highest point) 47-45, 46-94. (c) 605 ft (d) 845 ft 6. A rising gradient of 1 in 500 meets another rising gradient of 1 in 400 at a level of 264-40 ft, and a second gradient 600 ft long then meets a falling gradient of 1 in 600. The gradients are to be joined by two transition curves, each 400 ft long. Calculate the levels on the curves at 100 ft intervals. (L.U. Ans. 264-00, 264-21, 264*42, 264-65, 264-90, 265-15, 265-40, 265*61, 265-70, 265*69, 265-58) 7. A falling gradient of 4% meets a rising gradient of 5% at chainage 2450-0 ft and level 216-42 ft. At chainage 2350, the underside of a bridge has a level of 235-54 ft. The two gradients are to be joined by a vertical parabolic curve giving 14 ft clearance under the bridge. List the levels at 50 ft inter- vals along the curve. 626 SURVEYING PROBLEMS AND SOLUTIONS (L.U. Ans. 224-42, 222-70, 221-54, 220-95, 220-92, 221-45, 222-54, 224-20, 226-42) 8. The surface of a length of a propsed road of a rising gradient of 2% is followed by a falling gradient of 4% with the two gradients joined by a vertical parabolic summit curve 400 ft long. The two gradients produced meet at a reduced level of 95*00 ft O.D. Compute the reduced level of the curve at the ends, at 100 ft inter- vals, and at the highest point. What is the minimum distance at which a driver whose eye is 3 ft 9 in. above the road surface would be unable to see an obstruction 4 inches high ? (I.C.E. Ans. 91-00, 92*25, 92-00, 90*25, 87*00 ft A.O.D. highest point 92-33 ft AOD ; sight distance 290 ft) 9. An existing length of road consists of a rising gradient of 1 in 20, followed by a vertical parabolic summit curve 300 ft long, and then a falling gradient of 1 in 40. The curve joins both gradients tangent- ially and the reduced level of the highest point on the curve is 173*07 ft above datum. Visibility is to be improved over the stretch of road by replacing this curve with another parabolic curve 600 ft long. Find the depth of excavation required at the midpoint of the curve. Tabulate the reduced levels of points at 100 ft intervals on the new curve. What will be the minimum visibility on the new curve for a driver whose eyes are 4*0 ft above the road surface? (I.C.E. Ans. 2*81 ft; 160-57, 164-95, 168-08, 169-95, 170-61, 169-99, 168*07 ft A.O.D. ; minimum visibility 253 ft) 10 A vertical curve 400 ft long of the parabolic type is to join a fall- ing gradient of 1 in 200 to a rising gradient of 1 in 300. If the level of the intersection of the two gradients is 101*20 ft, give the levels at 50 ft intervals along the curve. If the headlamp of a car was 1*25 ft above the road surface, at what distance will the beam strike the road surface when the car is at the start of the curve ? Assume that the beam is horizontal when the car is on a level surface. (L.U. Ans. 102-20, 101-98, 101-80, 101-68, 101-62, 101-60, 101-64, 101*72, 101*86; 347ft) 11. A vertical parabolic curve 500 ft long connects an upward gradient of 1 in 100 to a downward gradient of 1 in 50. If the tangent point T, between the first gradient and the curve is taken as datum, calculate the levels of points at intervals of 100 ft along the curve, until it meets VERTICAL AND TRANSITION CURVES 627 the second gradient at T z . Calculate also the level of the summit giving the horizontal dis- tance of this point from T,. If an object 3 in. high is lying on the road between T, and T 2 at 10 ft from T 2 and a car is approaching from the direction of T,, calcu- late the position of the car when the driver first sees the object, if his eyes are 4 ft above the road surface. (L.U. Ans. 0-70, 0-80, 0*30, -0*80, -2-50, 0'84, 166-67 ft 33-6 ft from T,) 12. A parabolic vertical curve of length L is formed at a summit between an uphill gradient of a% and a downhill gradient of b%. As part of a road improvement, the uphill gradient is reduced to c % and the downhill gradient increased to d%, but as much as possible of the original curve is retained. Show that the length of the new vertical curve is L x (c + d) (L.U.) (a + b) 13. The algebraic difference in the gradient of a sag vertical curve L ft long is a ft/ ft. The headlamps of a car travelling along this curve are 2*5 ft above the road surface and their beams tilt upwards at an angle of 1° above the longitudinal axis of the car. Show that if s, the sight distance in feet, is less than L, then, L = ^ (L.U.) 5 + 0-035 s 12.6 Transition Curves 12.61 Superelevation (0) A vehicle of weight W (lbf) or (kgf) on a curve of radius r (ft) or (m), is travelling at a velocity v (ft/s) or (m/s) or V (mile/h). Fig. 12.16 shows the centrifugal force Wv 2 /gr which must be resisted by either (a) the rails in the case of a railway train, or (b) adhesion be- tween the road and the vehicle's tyres, unless superelevation is applied, when the forces along the plane are equalised. Then Wv 2 cosfl = Wsind (12.31) gr i.e. If 6 is small, tan0 = Z_ (12.32) gr rad = Z_ (known as the centripetal 8 r ratio) (12.33) 628 SURVEYING PROBLEMS AND SOLUTIONS cos 6 an W sin 6 Fig. 12.16 12.62 Cant (c) If d is the width of the track, then the cant c is given as c = dsind (12.34) dv 2 gr COS0 ~ ^l! (if e is small) gr (12.35) (12.36) N.B. c oc v 2 oc 1/r On railways c is usually limited to 6 in. with a 4 ft 8y2in. gauge. On roads tantf is usually limited to O'l. 12.63 Minimum curvature for standard velocity Without superelevation on roads, side slip will occur if the side thrust is greater than the adhesion, i.e. if Wv z /gr > /j,W, where jjl ~ the coefficient of adhesion, usually taken as 0'25. Thus the limiting radius r = Z_ If the velocity V is given in mile/h, (12.37) r = ft 15 M (12.38) VERTICAL AND TRANSITION CURVES 629 12.64 Length of transition Various criteria are suggested: (1) An arbitrary length of say 200 ft. (2) The total length of the curve divided into 3 equal parts, 1/3 each transition; 1/3 circular. (3) An arbitrary gradient of 1 in. in sft, e.g. lin. in 25 ft -the steepest gradient recommended for railways. (4) At a limited rate of change of radial acceleration (W.H. Short, 1908, fixed 1 ft/s 3 as a suitable value for passenger comfort.) (5) At an arbitrary time rate i.e. 1 to 2 in per second. N.B. (4) is the most widely adopted. 12.65 Radial acceleration The radial acceleration increases from zero at the start of the transition to v 2 /r at the join with the circular curve. The time taken t = — , where / = the length of v transition. .". the rate of gain of radial acceleration r ' v ,3 - 1- ft/s 3 rl (12.39) Thus the length of the curve I is given as l - 21. ar (12.40) If a is limited to 1 ft/s 3 then / = ll (ft) r (12.41) 3-155 V 3 (ft) (12.42) On sharp radius curves, the value of a would be too great, so the superelevation is limited and the speed must be reduced. From Eq. (12.32), ■ z tan = _ gr then v = y/(g.r tan<9) (12.43) v 3 but 1 = 1- ar 630 SURVEYING PROBLEMS AND SOLUTIONS / = (stan<9) 3/2 ^ (12.44) a i.e. when c is limited, / oc \Jr. For railways with a maximum superelevation of 6 in., sinfl = Ac 6 '™' = 0-1008 4 ft SVi in. 6 = 5° 47' tan# = 0-1013. Thus the maximum speed should be: v = VC 32 ^ x 0-1013 r) = 1-806 Vr ft/s (12.45) ^ 2y/r ft/s (12.46) (12.47) (12.48) 12.7 The Ideal Transition Curve If the centrifugal force F = Wv 2 /gr is to increase at a constant rate, it must vary with time and therefore, if the speed is constant, with distance. , i.e. r oc / oc gr I oc — i.e. rl = constant '= RL r where R = the radius of the circular curves L = the total length of the transition. In Fig. 12.17, 81 = rS<£ d<f> = -dl r but rl = RL = constant k dcf> = -j-dl If a = If t/s 3 , / r = Sy/r (2 V / r) 3 r If / and r are given in Guntei r chains, 66/ = 8V(66r) / * V chains Integrating, , / 2 k 2k VERTICAL AND TRANSITION CURVES 631 Fig. 12. 17 The ideal transition curve but </> = when / = c = = / = r 2RL WV0 where M = y/2RL (12.49) (12.50) This is the intrinsic equation of the clothoid, to which the cubic parabola and lemniscate are approximations often adopted when the deviation angle is small, Fig. 12.18. y (a) Clothoid (b) Lemniscate (c) Cubic parabola c "o 1 X Fig. 12. 18 (a) Clothoid I = My/</> (b) Lemniscate c 2 = a 2 sin 20 x 3 (c) Cubic parabola y = 6RX (y = — — cubic spiral ) V 6RL I 632 SURVEYING PROBLEMS AND SOLUTIONS 12.8 The Clothoid / = V(2/?L)V0 - l z /2RL where R = the minimum radius (i.e. the radius of the circular curve). As the variable angle <f> cannot be measured from one position, it is difficult to set out in this form. Fig. 12. 19 Clothoid 12.81 To find Cartesian co-ordinates dx dl = cos 2! 4! = 1 - ' + *' Integrating, [ l A Y ~ 5x2!(2 ■i 2! (2RL) 2 4! (2KL) 4 _ / 8 (2RL) 2 + 9x4!(2RL) 4 ' 1 - ^ +_£_ 5x2! 9x4! For <£max I = L then x ~ / I 40K 2 J Similarly, =¥. = sin <f> = $ - -$- + *£— dl 3! 5! r / 6 / ,0 + Integrating, y = / r 2RL 3!(2KL) 3 5!(2KL) 5 z 6 / ,0 3(2RL) 7x3!(2RL) 3 llx5\(2RLf <f> 3 - iTi _ _*!_ + _5^! 1 1.3 7x3! 11x5! J (12.51) (12.52) (12.53) (12.54) (12.55) (12.56) (12.57) (12.58) (12.59) VERTICAL AND TRANSITION CURVES 633 For <£ ma3t , 6R [ 56R 2 12.82 The tangential angle a Tana = X. = ± + _£! + x 3 105 but By substitution, a = tana - -|tan 3 a + Itan 5 a •3 5 a = ±- Ml 3 2835 (12.60) (12.61) (12.62) (12.63) i.e. a = & - k (a rapidly decreasing (12.64) Thus, if <p is small, quantity) a = 2- (12.65) Jenkins* shows that if <f> < 6°, no correction is required; and if < 20° no correction > 20" is required. 12.83 Amount of shift (s) The shift is the displacement of the circular curve from the tangent, i.e. DF, Fig. 12.20. T F G H Fig. 12. 20 Amount of shift ByEq.(12.59), PH = BF = y = l\^-— ^ + ...1 L 3 7x3! J DF = BF - BD = y - fl(l- cos<£) J 2 T Expanding cos^ and putting max = ^ = ^ DF = 24R * R. B.M.Jenkins, Curve Surveying (Macmillan). (12.66) (12.67) (12.68) 634 SURVEYING PROBLEMS AND SOLUTIONS 12.9 The Bernouilli Lemniscate Polar equation c 2 = a 2 sin 2a Identical to clothoid for deviation angles up to 60° (radius de- creases up to 135°), Fig. 12.21, c is maximum = a when a = 45° *\ \ *1 >l L 31 B \r=r« J^^y ^g£s£^ 0-V e A. \135 4 r, F N Fig. 12.21 Principles of the Bernouilli lemniscate Referring to Fig. 12.22, 2c 2 gives - c 2 = a 2 sin 2a 2c — = 2a 2 cos 2a da 1 dc 2a 2 cos 2a (12.69) cot 2a c da 2a 2 sin 2a If $ = angle T X PQ and P, and P 2 are 2 neighbouring points, cot0 = ^ (P 2 MP, =90°) P 2 lw 5c c8a i.e. Iff£ = cot0 c da :. = 2a <f> = + a = 3a as shown in the clothoid and the cubic parabola. (12.70) p,p 2 - a - cSa VERTICAL AND TRANSITION CURVES P 2 M c8a 635 sin 6 sin 6* sin 2a . dl c c c \ \Sr^ da sin 2a c 2 /a 2 ^jjz^r^ Now d/ = rd<£ = 3rda See c^- — ^^\^r da c ^s^ >^* Fig. 12.22 Thus a 2 = = 3rc (12.71) i.e., from Eq. (12.69) c 2 = = 3Rc sin 2a if the lemniscate approximates to the circle radius R c = = 3/? sin 2a (12.72) In Fig. 12.21 OF = OB + BF = r cos <f> + c sin a = r cos0 + 3r sin a sin 2a = r[cos3a + 3 sin a sin 2a] (12.73) T,F = T,N - FN = c cos a - r sin = 3r sin 2a cos a - r sin 3a = r[3 sin 2a cos a - sin 3a] (12.74) 12.91 Setting out using the lemniscate Using Eq. (12.72), c = 3R sin 2a Shift DF = OF - OD = R [cos 3a + 3 sin a sin 2a] - R = /?[cos3a + 3 sin a sin 2a - 1] (12.75) Equal values of a enable values of c to be computed, or equal chords c, 2c, 3c, etc. sin 2a = JL (12.76) 3R where a = $ 636 SURVEYING PROBLEMS AND SOLUTIONS If a is small, a" = 206265c 6R (12.77) <W - ^f^ (12.78) (The same value as with the cubic spiral) For offset values, y = c sin a (12.79) x = c cos a (12.80) If the chord lengths are required between adjacent points on the curve, the length of the chord c = y/{x 2 +y 2 ) (12.81) 12.10 The cubic parabola This is probably the most widely used in practice because of its simplicity. It is almost identical with the clothoid and lemniscate for deviation angles up to 12°. The radius of curvature reaches a mini- mum for deviation angles of 24° 06' and then increases. It is therefore not acceptable beyond this point. .3 Let y = A_ k (12.82) dy _ 3x 2 dx k (12.83) d y _ 6x dx 2 k (12.84) By the calculus the curvature (p) is given as 1 d 2 y/dx 2 p = — = il 3/2 (12.85) ( + \dx) If <f> is small, dy/dx is small and (dy/dx) is neglected. jl_ _ d y _ 6x 7 dx 2 ~ ~k~ (12.86) k = 6rx = 6RX at the end of (12.87) Equation of the cubic parabola e curve y = -^- (12.88) 6RX If the deviation angle <f> is small, x ~ /, X o- L. Equation of the cubic spiral y - <£ (1289) VERTICAL AND TRANSITION CURVES 637 N.B. This is the first term in the clothoid series. cf> ~ tan<£ _ dy _ 3x 2 dx 6RX x 2 2RX (12.90) a ^ tana _ y _ x 2 x 6RX (12.91) .\ a 3^ (12.92) as in the first term of the clothoid series. In Fig. 12.21, PB = /? sin <£ ^ /?<£ = X/2 (12.93) i.e. the shift bisects the length. DB = R(l-cos<f>) (12.94) Shift DF = y-/?(l-cos<£) (12.95) X a on _:_, ' 6RX - 2R sin* - X 2 2R<f> 2 6/? 4 x 2 6/? x 2 8/? X 2 24R L 2 24/? the first term in the Clothoid series As E is on transition and TF = FH = kx, EF %' X 3 X 2 (12.96) 6RX 48/?X \DF (12.97) 48/? i.e. the transition bisects the shift. 12.11 The Insertion of Transition Curves The insertion of transition curves into the existing alignment of straights is done by one of the following alternatives. (1) The radius of the existing circular curve is reduced by the amount of 'shift', Fig. 12.23. The centre is retained. /?, - R 2 = shift (5) = -L_ 2 24/? T'T if 1 1 * 1 a> 21-, 638 SURVEYING PROBLEMS AND SOLUTIONS Fig. 12.23 (2) The radius and centre are retained and the tangents are moved outwards to allow transition, Fig. 12.24. (N.B. Part of the original curve is retained.) /,/ 2 = 0I 2 - 0/, = R + S R cos — COS — 2 2 cos Fig. 12.24 VERTICAL AND TRANSITION CURVES 639 (3) The radius of the curve is retained, but the centre is moved away from the intersection point, Fig. 12.25. o,o 2 = shift cos u Fig. 12.25 (4) Tangent, radius and part of the existing curve are retained, but a compound circular curve is introduced to allow shift, Fig. 12.26. Fig. 12.26 (5) A combination of any of these forms. 640 SURVEYING PROBLEMS AND SOLUTIONS 12.12 Setting-out Processes Location of tangent points (Fig. 12.27) TyP, Transition curve PiP a Circular curve %Ti — Transition curve Fig. 12.27 Setting-out processes By Eq. (12.96), FD shift (s) = ±L 24R FI = (R + s) tan iA FT ^ H Tangent length I,/ = T 2 I = ^L + (/? + s) tan 5A (12.98) Setting out (Fig. 12.27) (1) Produce straights (if possible) to meet at /. Measure A. (2) Measure tangent lengths /T, = IT 2 to locate T, and T 2 . (3) Set out transition from T, . Two methods are possible, (a) by offsets from the tangent, (b) by tangential angles. For either method, accuracy is reduced when / > 0-4R i.e. <f> > 12° Method (b) is more accurate, even assuming chord = arc. VERTICAL AND TRANSITION CURVES 641 (4) Offsets from the tangent (Fig. 12.28) From Eq. (12.88), ~3 y, = 6RL IX X 2 = «X < 9 X g = ox ^ then y 2 = 2 3 y, = 8y, y 3 = 3 3 y, = 27y, Fig. 12.28 Offsets from the tangent (5) Tangential angles a (Fig. 12.29) Fig. 12.29 Deflection angles from the tangent From Eq. (12.89) tana If a is small, then if x ^ c a = 6RL 206 265 x 2 6RL 206 265 c 2 6RL (12.99) (12.100) For a" c^ L max — a' ^ 573 c 2 /RL OL_ = 206 265 c 6R a' = 573 c/R (12.101) (12.102) 642 SURVEYING PROBLEMS AND SOLUTIONS 0m«* = 3a m = 206 265c/2K (12.103) <f>' m = 1719 c/R (12.104) (6) Check on P, (Join of transition to circular curve) N,P, = N 2 P 2 = ^ (12.105) (7) Move theodolite to P,. Set out circular curve by offsets or deflection angles from the tan- gent QPZ. N.B. Angle T,P,Q, = 6 = 2a = §0 m (12.106) Checfc ZP,P 2 = /8 = | A - <£ m (12.107) P,P 2 = 2R sinjS (12.108) Check N 2 P 2 = N,P, = ^ Eq. (12.105) (8) Move theodolite to T 2 and set out the transition backwards towards P 2 as in (3). N.B. The use of metric units does not make any difference to the solu- tion, providing these are compatible, i.e. v - m/s L and R - m a - ft/s 3 converted to m/s 3 (e.g. lft/s 3 = 0-305 m/s 3 ) Example 12.6. A circular curve of 2000 ft radius deflects through an angle of 40° 30'. This curve is to be replaced by one of smaller radius so as to admit transitions 350 ft long at each end. The deviation of the new curve from the old at their midpoint is 1*5 ft towards the intersect tion point. Determine the amended radius assuming the shift can be calculated with sufficient accuracy on the old radius. Calculate the length of the track to be lifted and the new track to be laid. (L.U.) By (12.94) Shift ,, . JL . ^|5l_ .. 2 . 55 ft . Tangent length of circular curve (7J /) = 2000 tan40°30^- = 737*84 ft. In triangle 0,/T, ftl . -JSSJL-, . 2131 . 8ft X,/ = 2131*8 - 2000 = 131 -8 ft new value XJ = 131*8 - 1*5 = 130*3 ft VERTICAL AND TRANSITION CURVES 643 Not to scale In triangle OJF A - 40* 30' = cos 20° 15' = 0*93 819 i.e. By (12.103) By (12.107) R + 130-3 K + 2-55 = (R + 130-3)0-93819 •• R = Trim 5 ■ i2*Lift 206265x350 1fi ,, Q „ *- = 3a = 2 x 1936-5 = 18639 = 5° 10' 39" = iA - <£ max = 20° 15' - 5° 10' 39" = 15° 04' 21" To find length of track to be lifted (T, T 2 ) Length of circular curve = 2000 x 40°30' rad = 1413-72 ft 2 + (R + s)tany 2000 Tangent length new shift T / = 4 3 S * ' S ' * 1936-5 - 2-SSx,^ = 2-64 1936*5 T 3 1 = 175*0 + (1936-5 + 2*64) tan 20° 15' = 175-0 + 715-39 = 890-39 ft but 7;/ = 737-84 ft .-. TJ 3 = 152-55 ft therefore total length of track to be lifted = length of arc + 2 x T, T 3 = 1413-72 + 305-10 = 1718-8 ft 644 SURVEYING PROBLEMS AND SOLUTIONS To find the length of track to be laid = 2 x transition curve (T 3 fj ) + circular arc (f| P z ) = 2 x 350 + 2 x 1936-5/3 = 700 + 3873-0 x 0*26306 = 700 + 1018-83 = 1718-8 ft. 12.13 Transition Curves Applied to Compound Curves In this case, the transition would be applied at the entry and exit (i.e. at T, and T 2 (Fig. 12.31) The amount of superelevation cannot be designed to conform to both circular arcs and the design speed must relate to the smaller radius. Fig. 12.31 Transition curves applied to compound curves If the two curves are to be con- nected by a transition curve of length /, the shortest distance c (Fig. 12.32) between the two circu- lar curves is given by Glover* as c '^(w,~w) (12109 > The length of the transition L, is bisected at Q by this shift c and the shift is bisected by the transition. ^-- * Fig. 12.32 ♦Transition curves for railways, Proc. Inst. Civ. Eng., Vol. 140. VERTICAL AND TRANSITION CURVES 645 If transitions are applied to reverse curves, the radii must be re- duced to allow the transition curves to be introduced, Fig. 12.33. Fig. 12.33 Transitions applied to reverse curves /,/ 2 = /,T 2 + T Z I 2 = (R, +s,) tan§A, + §L, + \L 2 + (R 2 + s 2 ) tan §A 2 (12.110) If L = yJR (based on Gunter chains), L 2 24R R 24/? — chains (12.111) 24 Shift s, = s 2 •'• A/ 2 = (*?, + ^) tan |A, + yR, + \y/R 2 + (R 2 + ^) tan \ A 2 (12.112) If R, = R 2 , then Uh = (/? + 2^)(tan|A,+tan|A 2 ) + y/R (12.113) This may be solved as a quadratic in ^R Example 12.7. Two railway lines have straights which are deflected through 70°. The circular radius is to be 1500 feet with a maximum superelevation of 5 in. The gradient of the line is to be 1 in. in 1 chain (Gunter) . Calculate the distance from the beginning of the transition to the intersection point (i.e. tangent length), the lengths of the separate portions of the curve and sufficient data for setting out the curve by 646 SURVEYING PROBLEMS AND SOLUTIONS offsets from the tangent and by the method of tangential angles. Fig. 12.34 Gradient is 1 in. in 66 ft, therefore for the total superelevation it is 5in. Shift = 66 ft = 330 ft 330 2 24R 24 x 1500 3-03 ft <*. T.I = 1503-03 tan 35° + M 2 = 1217-43 ft tan" 1 A- = tan" 1 -^ = 6° 16' 37" 2R 3000 j8 = 35°- 6° 16' 37" .= 28° 43' 23" Length of circular curve = 1500 x 57°26'46;' ad = 1503*93 ft Offsets from tangents y = 6RL Let the curve be subdivided into 5 equal parts. (1 chain each.) 66 3 y^ = 6 x 1500 x 330 y 2 = 0-0968 x 2 Z y 3 = 0-0968 x 3 J y 4 = 0-0968 x 4 ; = 0-096 8 ft = 0-77 ft = 2-61 ft = 6-19 ft VERTICAL AND TRANSITION CURVES 647 y_ = 0-0968 x 5 3 = 12*10 ft Tangential angles, based on 1 chain chords. a, = a 2 a, a A a. 206 265 c 2 206 265 x 66' 6RL = 4a, = 9a, = 16a, = 25a, = 3^ ~ 6x1500x330 = 1210" = 2722-5" = 4840" = 7562-5 = 2° 05' 32" error = 3025" (05' 03") = 20' 10" = 45' 23" = 80' 40" = 126 '03" = 125' 32" 31" Example 12.8. An existing circular curve of 1500 ft radius is to be improved by sharpening the ends to 1300 ft radius and inserting a trans- ition curve 400 ft long at each straight. Using the cubic parabola type, calculate: (a) the length of curve to be taken up, (b) the movement of the tangent points, (c) the offsets for the quarter points of the transition curves. (L.U.) V y\2od 02 "^ -^^ K ^ *^\\ \ \l300' ^w B -_ D >* ~-- ■ Fig. 12.35 This is Case (4) in Section 12.11. In Fig. 12.35, PN = y = 6R 400' 1300 = 20-51 ft 648 SURVEYING PROBLEMS AND SOLUTIONS cf> m = tan-'i^ 2R 400 = Lelll ' 2 : In triangle 2 BP, = tan-' Z^y = 8°48' 2 x 1300 2 B = 1300cos8°48' = 1284-66 ft BP = FN = 1300 sin 8° 48' = 198*90 ft 0,V = 0,7, - WT, -WV= OJi - PN - 2 B = 1500 - 20-51 - 1284-66 = 194-83 ft In triangle 0,0 2 V , X = cos" 1 194 ' 83 = 13° 05' 200-0 V0 2 - 200sinl3°05' =45-28 ft (a) Length of curve taken up = 1500 x 13°05; ad = 342-5 ft. (b) Movement of the tangent points, i.e. distance T y T 2 : FN = T 2 F = 198-90 ft T y F = V0 2 = 45-28 ft T,T 2 = T 2 F - T X F = 153-62 ft i.e. along straight from T, . (c) Offsets to transition: I 3 y = r v i = 6RL 100 3 = 0-320 5 ft = 0-32 ft 6 x 1300 x 400 y 2 = 2 3 y, = 8y, = 2-56 ft y 3 = 3 3 y, = 27y, = 8-66 ft y 4 = 4 3 y, = 64y, = 20*51 ft check Example 12.9. AB, BC and CD are three straights. The length of BC is 40 Gunter chains. BC deflects 60° right from AB and CD 45° left from BC. Find the radius r for two equal circular curves, each with transition curves of length y/r at both ends to connect AB and CD. BC is to be the common tangent without intermediate straight. Find also the total length of curve. (L.U.) VERTICAL AND TRANSITION CURVES 649 By Eq. (12.112), 40-0 = (r +2^) (tan 30° + tan 22° 30') + V = (r+ 2 ^-)(0-5774+ 0-414 2) + yjr i.e. 0-991 6r + y/r - 39-96 = Solving this quadratic equation in yV, i.e. let / = r, 1=1 = -1 ± V(l + 4 x 0-9916x39-96) V 2 x 0-9916 / = 5-864 chains r = 34-39 chains 206 265 L _ 206 265^ _ 103132 x 5-864 9max 2 R 2r 68-78 = 8793" = 2° 26' 33" /8, = 30 - 2° 26' 33" = 27° 33' 27" Length of circular arc, A x = 2 x 34-39 x 27° 33' 27 r " ad = 68-78 x 0-48096 = 33-08 ch ft, = 22° 30' - 2° 26 ' 33" = 19° 33' 27" A 2 = 2 x 34-39 x 19° 33' 27" = 68-78 x 0-34133 = 23-48 ch Total length - A, + A z + 2L = 33-08 + 23-48 + 11*73 = 68-29 chains Exercises 12(b) 14. A road curve of 2000 ft radius is to be connected to two straights by means of transition curves of the cubic parabola type at each end. The maximum speed on this part of the road is to be 70 mile/h and the rate of change of radial acceleration is 1 ft/s 3 . The angle of intersec- tion of the two straights is 50° and the chainage of the intersection point is 5872-84 ft. Calculate: (a) the length of each transition curve, (b) the shift of the circular arc, (c) the chainage at the beginning and the end of the composite curve, (d) the value of the first two deflection angles for setting out the 650 SURVEYING PROBLEMS AND SOLUTIONS first two pegs of the transition curve from the first tangent point as- suming that the pegs are set out at 50 ft intervals. (I.C.E. Ans. (a) 542-1 ft (b) 6-10 ft (c) 4666-33 ft; 6955-93 ft (d) 44", 3' 39") 15. Two tangents which intersect at an angle of 41° 40' are to be connected by a circular curve of 3000 ft radius with a transition curve at each end. The chainage of the intersection point is 2784 + 26. The transition curves are to be of the cubic parabolic type, designed for a maximum speed of 60 mile/h and a rate of change of radial acceleration is not to exceed 1 ft/s 3 . Find the chainage of the beginning and end of the first transition curve and draw up a table of deflection angles for setting out the curve in 50 ft chord lengths, chainage running continuously through the tan- gent point (I.C.E. Ans. 2771 + 70-5; 2773 + 97-7; 40"; 5'08"; 13' 54"; 26' 42"; 43' 18" ) 16. The limiting speed around a circular curve of 2000 ft radius calls for a superelevation of 1/24 across the 30 ft carriageway. Adopting the Ministry of Transport's recommendation of a rate of 1 in 200 for the application of superelevation along the transition curve leading from the straight to the circular curve, calculate the tangential angles for setting out of the transition curve with pegs at 50 ft intervals from the tangent with the straight. (I.C.E. Ans. 02' 52"; 11' 28"; 25' 48"; 45' 52"; 1°11'40") 17. Two straights of a proposed length of railway track are to be joined by a circular curve of 2200 ft radius with cubic parabolic transi- tions 220 ft long at entry, and exit. The deflection angle between the two straights is 22° 38' and the chainage of the intersection point on the first straight produced is 2553-0 ft. Determine the chainages at the ends of both transitions and the information required in the field for setting out the midpoint and end of the first transition curve. If the transition curve is designed to give a rate of change of radial acceleration of 1 ft/s 3 , what will be the superelevation of the outer rail at the midpoint of the transition, if the distance between the centres of the rails is 4 ft 11 in. ? (I.C.E. Ans. 2002-6 ft; 2222-6 ft; 2871-6 ft; 3091-6 ft; Offsets 0-46 ft and 3-67 ft; 2-2 in.) 18. A transition curve of the cubic parabola type is to be set out from a straight centre line. It must pass through a point which is 20 ft away from the straight, measured at right-angles from a point on the straight produced, 200 ft from the start of the curve. Tabulate the data for setting out a 400 ft length of curve at 50 ft intervals. VERTICAL AND TRANSITION CURVES 651 Calculate the rate of change of radial acceleration for a speed of 30 mile/h. (L.U. Ans. 0°20'20", 1°26'00", 3° 13' 20", 5° 42' 40", 8°52'50", 12° 40' 50", 17° 01' 40", 21° 48' 05"; 1-28 ft/s 3 ) 19. Two straight portions of a railway line, intersecting at an angle of 155°, are to be connected by two cubic parabolic transition curves, each 250 ft long and a circular arc of 1000 ft radius. Calculate the necessary data for setting out the curve using chords 50 ft long. (R.I.C.S./M Ans. shift 2-61 ft; tangent length 4647-9 ft; 0°05'40", 0°23'00", 0°51'30", 1°31'40", 2° 23' 10") 20. Two straights of a railway with 4' SVi gauge, intersect at an angle of 135°. They are to be connected by a curve of 12 chains radius with cubic parabolic transitions at either end. The curve is to be designed for a maximum speed of 35 mile/h with a rate of gain of radial acceleration of 1 ft/s 3 . Calculate (a) the required length of transition, (b) the maximum super elevation of the outer rail, (c) the amount of shift required for the transition, and (d) the lengths of the tangent points from the intersec- tion of the straights. (R.I.C.S./M Ans. 170-7 ft, 5-8 in., 1-53 ft, 2001-1 ft) 21. Two railway straights, having an intersection (deviation) angle of 14° 02' 40", are to be connected by a circular curve of radius 2000 ft with spiral transitions at each end. (a) Calculate the superelevation for equilibrium on the circular arc, if the design speed is 45 mile/h, g = 32-2 ft/s 2 and the effective gauge between rails = 5 ft and thence, (b) if this super elevation is introduced with a gradient of 1 in 600, what is the length of each transition and of the circular curve. (c) Hence, given the point of intersection of the straights, compute all data required for setting out one of the spirals by means of deflec- tion angles and 50 ft chords. (R.I.C.S./L Ans. 4 in., 200ft, 3' 35", 14' 19", 32' 13", 57' 17") 22. (a) Calculate the setting out data for a circular curve, radius 500 ft joining two straights with a deviation angle of 30°00'00". (b) Show that a curve having a polar deflection angle equal to one third of its tangent deflection angle is a lemniscate. For the lemniscate, the ideal transition curve relationship between length of curve and radius of curvature does not hold. Show why this is not usually important. (N.U.) 652 SURVEYING PROBLEMS AND SOLUTIONS 23. The curve connecting two straights is to be wholly transitional without intermediate circular arc, and the junction of the two transi- tions is to be 16 ft from the intersection point of the straights which deflects through an angle of 18°. Calculate the tangent distances and the minimum radius of curva- ture. If the superelevation is limited to 1 vertical to 16 horizontal, determine the correct velocity for the curve and the rate of gain of radial acceleration. (L.U. Ans. 304-2 ft; 958-3 ft; 30mile/h; 0-292 ft/s 3 ) 24. The superelevation on a road 50 ft wide is to be 3 ft. Calculate the radius for a design speed of 40 mile/h and then give the data for setting out the curve if the two straights have a deflection angle of 30°. Transition curves 300 ft long will be applied at each end, but the data for setting out of these is not required. (L.U. Ans. Rad 1781*5 ft; tangent length 627'9 ft; shift 2-10 ft; Circular arc 632-8 ft) 25. Two straights of a road 20 ft wide intersect at a through chain- age of 8765-9 ft, the deflection angle being 44° 24'. The straights are to be connected by a circular arc of radius 900 ft with cubic parabolic transitions at entry and exit. The curve is to be designed for a speed of 45 mile/h, with a rate of gain of radial acceleration of 2 ft/s 3 . De- termine the required length of the transition and the maximum super- elevation of the outer kerb. Tabulate all the necessary data for setting out the first transition with pegs at every 50 ft of through chainage. (L.U. Ans. L = 159-7 ft; c = 3-0 ft; Chainage T, - 8318-3 ft; Offsets 0-04, 0-63, 2-65, 4-72 ft) 26. Assuming an equation A = /(<£) where A and are the intrinsic co-ordinates of any point on a transition spiral, prove that A 2 = 2RL<t>, where R = minimum radius of curvature; L = length of spiral, <p -.= the spiral angle. A curve on a trunk road is to be transitional throughout with a total deviation of 52° 24'. The design speed is to be 60 mile/h, the maximum centripetal ratio 0-25, and the rate of change of radial acceleration 1 ft/s 3 . Calculate (1) the length of each spiral, (2) the minimum radius of curvature, (3) the tangent distance. (L.U. Ans. L = 879-8 ft; R = 962 ft; T,/ = 926-6 ft) 27. A suburban road 30 ft wide, designed for a maximum speed of 40 mile/h is to deflect through 38° 14' with a radius of 1600 ft. A cubic parabola transition is required, with a rate of gain of radial accelera- tion of 1 ft/s 3 . Calculate (a) the maximum superelevation of the outer kerb, VERTICAL AND TRANSITION CURVES 653 (b) the length of the transition, (c) the chainage of the tangent points if the forward chainage of the intersection point is 5829*60 ft, (d) the chainages of the junctions of the transition and circular arcs. (N.R.C.T. Ans. (a) 2-0ft (b) 126-22 ft (c) 5211-77; 6405-69 ft (d) 5337-99; 6279-47 ft) 28. The co-ordinates of three points, K,L and M are as follows: Point North (ft) East (ft) K 700 867 L 700 1856 M 1672 2031 These points define the direction of two railway straights JK(M) and LM, which are to be connected by a reverse curve formed by circu- lar arcs of equal radius. The circular arcs are to be linked together and to the straights by easement curves of length (in Gunter's chains) equal to y/R where R is the radius of the circular arcs in chains. Calculate the radius of the circular arcs. (L.U. Ans. 9-87 chains) 29. A road 30 ft wide is to turn through an angle of 26° 24' with a centre line radius of 600 ft, the forward chainage of the intersection point being 3640-6 ft. A transition curve is to be used at each end of the circular curve of such a length that the rate of gain of radial ac- celeration is 1 ft/s 3 when the speed is 30 mile/h. Find the length of the transition curve, the banking of the road for this speed, the chain- age of the beginning of the combined curve, and the angle to turn off these for the peg at 3500 ft. (L.U. Ans. 142 ft; 2'99ft; 3428-6 ft; 0°34'20") 30. A road curve of 2000 ft radius is to be connected by two straights by means of transition curves of the cubic parabola type at each end. The maximum speed on this part of the road is to be 70 mile/h and the rate of change of radial acceleration is 1 ft/s 3 . The angle of intersec- tion of the two straights is 50° and the chainage at the intersection point is 5872-84 ft. Calculate: (a) the length of each transition curve, (b) the shift of the circular arc, (c) the chainage at the beginning and end of the composite curve, (d) the value of the first two deflection angles for setting out the first two pegs of the transition curve from the first tangent point, as- suming that the pegs are set out at 50 ft intervals. (I.C.E. Ans. (a) 541ft, (b) 6-10 ft (c) 4666-95 ft; 6953-25 ft (d) 35"; 3' 40") 654 SURVEYING PROBLEMS AND SOLUTIONS 31. A circular curve of radius 700 ft and length 410*70 ft connects two straights of railway track. In order that the track may be modern- ised to allow for the passage of faster traffic and induce less track wear, the whole curve and certain lengths of the connecting straights are to be removed and replaced by a new circular curve of radius 2500 ft, with transitions of the cubic parabola type at entry and exit. Given that the maximum speed of the traffic on the new curve is to be 60 mile/h, and the rate of change of radial acceleration is not to exceed 0'90 ft/s 3 , determine: (a) the length of the new composite curve, (b) the length of the straight track to be removed (c) the necessary superelevation of the track on the circular curve, the gauge of the track being 4 ft 8^ in. (I.C.E. Ans. (a) 1769-7 ft; (b) 2 -695-7 ft; (c) 5-42in) Bibliography CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) BANISTER, A. and Raymond, s., Surveying (Pitman) JAMESON, A.H., Advanced Surveying (Pitman) JENKINS, R.B.M., Curve Surveying (Macmillan) searles, w.H. and IVES, H.C., ed. by p. KIRSAM, Field Engineering, 22nd Ed. (John Wiley) HIGGINS, A.L., Higher Surveying (Macmillan) SHORTT, W.H., A Practical Method for Improvement of Existing Railway Curves, Proc. Inst. Civ. Eng., Vol. 176 ROYAL-DAWSON, F.G., Elements of Curve Design for Road, Railway and Racing Track (Spon,1932) ROYAL-DAWSON, F.G., Road Curves for Safe Modern Traffic (Spon, 1936) PERROTT, S.W. and BADGER, F.E.G., The Practice of Railway Survey- ing and Permanent Way Work (Edward Arnold, 1920) SHARMA, R.C. and SHARMA, S.K., Principles and Practice of Highway Engineering (Asia Publishing House) ^ problems & Solutions $8 Snenherd SBN: 7131 3198 5 ARNOLD